Let $a$ and $b$ be positive numbers such that $a+b=1$. Prove that: $$a^{4b^2}+b^{4a^2}\leq1$$

I think this inequality is very interesting because the equality "occurs" for $a=b=\frac{1}{2}$ and also for $a\rightarrow0$ and $b\rightarrow1$.

I tried to work with a function of one variable, but the derivative is not easy.

I also don't get something solvable by Taylor series.

  • 25,185
  • 9
  • 47
  • 115
Michael Rozenberg
  • 185,235
  • 30
  • 150
  • 262
  • 6
    A typical start should be $\,a=\sin^2{\theta}\,$: $$ \left(\sin{\theta}\right)^{8\cos^4{\theta}}+\left(\cos{\theta}\right)^{8\sin^4{\theta}}\le1\quad\colon\,\theta\in\left[0,\frac{\pi}{2}\right] $$ – Hazem Orabi Mar 07 '17 at 21:28
  • 1
    Intriguing inequality, where does it come from ? – Gabriel Romon Mar 12 '17 at 21:31
  • Looking at this yields the arguably tighter problem of $x^2+y^2 = 1 \implies x^{4y^2} + y^{4x^2} \le 1$ when $x,y >0$ – Brevan Ellefsen Mar 13 '17 at 04:57
  • 1
    To clarify what I meant in that last comment: We can sandwich the line $x+y=1$ in our interval between the curves $x^2+y^2 = 1$ and $x^2+y^2=\frac{1}{2}$, both of which satisfy the inequality. With a bit of work we might be able to massage a proof out of this. – Brevan Ellefsen Mar 13 '17 at 06:05
  • Using this method, all we have to solve is $$u+v = 1 \implies u^{2v}+v^{2u}\le1 \tag{1}$$ and an extremely similar second problem, then show that $x+y=1$ is strictly sandwiched between the two. I can post a solution using this method when I have time unless someone spots a blatant error with it. In the mean time, if anyone has a clever idea for solving $(1)$ please share... I'm thinking about just brute forcing it, which would be simpler to do than the original problems but still not very fun. – Brevan Ellefsen Mar 13 '17 at 06:28
  • One method might be to parametrize the same way we would with $x^y+y^x$ and seek an answer in terms of the $W$ function. Inequalities/Derivatives would then be easier.... I think? – Brevan Ellefsen Mar 13 '17 at 06:33
  • I find that with a=2 and b=-1, the condition is violated. Perhaps you need to specify that a and b are in the interval [0, 1]. – richard1941 Mar 14 '17 at 22:27
  • One could also define $f(x,y) = x^{4y^2} + y^{4x^2}$ and find its critical point with the constraint $x+y=1$ and prove that it's a maximum and it is equal to $1$. – embedded_dev Mar 16 '17 at 03:08
  • The answer to my $(1)$ above can be found on MathOverflow [here](http://mathoverflow.net/q/17189/83537), which explains that $(1)$ is non-trivial. Using that link as a reference proof I can probably show this inequality holds later today. I might also try a proof using Lagrange Multipliers, since that seems to be sorely lacking – Brevan Ellefsen Mar 16 '17 at 18:53
  • @MichaelRozenberg May I ask where did you get it? Interesting place it must be where you can find such problems!:) – Agile_Eagle Mar 17 '17 at 04:36
  • @MichaelRozenberg I have finished. Please now attack my post as viciously and rigorously as possible until any potential issues are ironed out. – Brevan Ellefsen Mar 19 '17 at 08:19
  • You mention [here](https://math.stackexchange.com/questions/2176571/if-ab-1-so-a4b2b4a2-leq1#comment6430362_3120958) that you have a proof now– why don't you post it as an answer? – Martin R Jul 17 '19 at 09:23
  • 1
    @MichaelRozenberg I have a proof wich is long but interesting if you want I can post it next week end. – Erik Satie Feb 13 '20 at 17:19

7 Answers7


We define $f(x,y)=x^{4y^2}+y^{4x^2}$.

This is my plan to solve the problem:

  • Since $x+y=1$, we replace $y$ by $1-x$.
  • We make a new function: $g(x)=x^{4(1-x)^2}+(1-x)^{4x^2}$
  • Therefore, we must find the maximum on the range $x \in [0,1]$ of $g$ so that we can see the maximum is less than or equal to $1$.

This will be troublesome:


Set $g_{1}(x) = x^{4(1-x)^2}$ and $g_{2}(x) = (1-x)^{4x^2}$. Therefore, we can break it up like so:

$$g'(x) = g_{1}'(x)+g_{2}'(x)$$ $$g_{1}'(x)=g_{1}', g_{2}'(x)=g_{2}'$$ $$\ln(g_{1})=\ln \left(x^{4(1-x)^2}\right)$$ $$\ln(g_{1})={4(1-x)^2} \cdot \ln \left(x\right)$$ $$\frac{g_{1}'}{g_{1}}= 4 \cdot \left((1-x)^2 \right)' \cdot \ln(x)+\frac{4(1-x)^2}{x}$$ $$\frac{g_{1}'}{g_{1}}= 4 \cdot -2 \cdot (1-x) \cdot \ln(x)+\frac{4(1-x)^2}{x}$$ $$\frac{g_{1}'}{g_{1}}= 8(x-1)\ln(x)+\frac{4x^2-8x+4}{x}$$ $$\frac{g_{1}'}{g_{1}}= 8(x-1)\ln(x)+4x-8+\frac{4}{x}$$ $$g_{1}'= x^{4(1-x)^2} \cdot \left(8(x-1)\ln(x)+4x-8+\frac{4}{x}\right)$$

Alright. Deep breath. Let's keep going.

$$\ln(g_{2})=4x^2\ln(1-x)$$ $$\frac{g_{2}'}{g_{2}}=8x\ln(x-1)+\frac{4x^2}{x-1}$$

$$g_{2}'= (1-x)^{4x^2}\left(8x\ln(x-1)+\frac{4x^2}{x-1}\right)$$

$$g_{1}'= x^{4(1-x)^2} \cdot \left(8(x-1)\ln(x)+4x-8+\frac{4}{x}\right)$$

$$g'(x)=x^{4(1-x)^2} \cdot \left(8(x-1)\ln(x)+4x-8+\frac{4}{x}\right) + (1-x)^{4x^2}\left(8x\ln(1-x)+\frac{4x^2}{x-1}\right)$$

The maximum appears (according to the closed interval method), either at:

$$g(0)=1$$ $$g(1)=1$$

Or at the $x$-value(s) of the solution of:

$$0=x^{4(1-x)^2} \cdot \left(8(x-1)\ln(x)+4x-8+\frac{4}{x}\right) + (1-x)^{4x^2}\left(8x\ln(1-x)+\frac{4x^2}{x-1}\right)$$

Therefore, if we set $x_{1}$, $x_{2}$, $x_{3}$ ... to be the solutions to the equation above in the interval $x_{n} \in [0,1]$, we have reduced the problem to proving that:

$$g(x_{1}),g(x_{2}), g(x_{3})... \leq 1$$

Through some graphing of $g(x)$, we see that there exists $x_{1}$, $x_{2}$, and $x_{3}$, where $x_{2}$ is $0.5$ and the others are not easily calculatable or are irrational.

It can easily be seen that $g'(0.5) = 0$ and that $g(0.5)=1$ (a maximum of the function). Since we now have proof that $g(x_{2}) \leq 1$ and we see that there does not exist an $x_{n}$ s.t. $n>3$ and $g'(x_{n})=0$, we can reduce our previous problem to:

Prove that:

$$g(x_{1}), g(x_{3}) \leq 1$$

Through Newton's Method, we obtain approximations of $x_{1}$ and $x_{3}$ accurate to 10 decimal places. We state them below:

$$x_{1} \approx 0.281731964017$$ $$x_{3} \approx 0.718268035983$$

Note that:

$$g'(x_{1}) \approx g'(0.281731964017)=7.349676423 \cdot 10^{-14}$$ $$g(x_{1}) \approx g(0.281731964017)=0.973494223187$$

We now have that $g(x_{1})$ is a minimum of the function and that $g(x_{1}) \leq 1$


$$g'(x_{3}) \approx g'(0.718268035983)=-7.349676423 \cdot 10^{-14}$$ $$g(x_{3}) \approx g(0.718268035983)=0.973494223187$$

We now have that $g(x_{1})$ is also a minimum of the function and that $g(x_{1}) \leq 1$

We now have that:

$$g(x_{1}), g(x_{2}), g(x_{3}) \leq 1$$


I took a very head-on brute-force approach to the problem, but I am happy with the rigorousness of the result and the final proof. We also now have the minimums of the function, which if anyone is curious, is $\approx 0.973494223187$

Maximilian Janisch
  • 12,071
  • 2
  • 15
  • 41
Robin Aldabanx
  • 1,853
  • 1
  • 10
  • 19
  • And wow Brevan Ellefsen's comment that is way easier to prove. – Robin Aldabanx Mar 13 '17 at 05:41
  • 3
    hmm. Basically brute forcing the problem. Works well enough, but based on the beautiful visual in @masnuhuey's answer I feel like we should be able to do this so much more simply. If only we could rotate our main function 45 degrees counter-clockwise... our line $x+y=1$ becomes $y =\frac{1}{\sqrt{2}}$ but I'm not sure how to parametrize our inequality. – Brevan Ellefsen Mar 13 '17 at 05:59
  • Btw, you mentioned my comment without tagging me with the @ symbol so I just noticed it by chance. I don't immediately see how to solve the problem $x^2+y^2=1$, but it felt like a cleaner start because we could bound with a cleaner curve. Note that if we solve the problem for $x^2+y^2=\frac{1}{2}$ as well we can probably pull out a cleaner proof, as $x+y=1$ is sandwiched between the two – Brevan Ellefsen Mar 13 '17 at 06:04
  • what is the "closed interval method"? – Ewan Delanoy Mar 13 '17 at 06:17
  • (One of) the interior maximum is at the middle of the interval, with the value 1 – Bananach Mar 13 '17 at 06:42
  • @EwanDelanoy it states that for a one-variable function $f(x)$ defined, continuous, and differentiable on a closed interval $[a, b]$, the maximum and minimum of $f(x)$ either appear at $a$, $b$, or when $f'(x)=0$ – Robin Aldabanx Mar 13 '17 at 23:27
  • 1
    You get the same root equation using a Lagrange multiplier, although the algebra is a bit simpler. I don't see an obvious analytical solution for the root. – Brady Gilg Mar 13 '17 at 23:30
  • @BradyGilg That's why I used Newton's method. Yes, that is true, although I don't feel perfectly comfortable using Lagrange multipliers yet. I will admit that the constraint $x+y=1$ makes it perfect for that Lagrange multipliers technique. I probably could've simplified this tremendously using that technique. – Robin Aldabanx Mar 13 '17 at 23:33
  • Thank you, I will fix it right away. – Robin Aldabanx Mar 17 '17 at 05:21
  • It has been fixed, thank you. – Robin Aldabanx Mar 17 '17 at 05:39
  • 1
    @Robin Aldabanx Thank you for your trying, but for me it's not proof. I am sorry. I am looking for an human proof, which we can use during the competition. My congratulations for $+200$. – Michael Rozenberg Mar 17 '17 at 17:35
  • @MichaelRozenberg I have actually pretty much already written up a human proof. I couldn't get it in time for the bounty (not that I care too much about rep anyway, as long as the math gets done!). I'll try to post it soon. I rely on the existing result $a+b=1 \implies a^{2b}+b^{2a} \le 1$, which is relatively well known in the literature. I don't think any trivial solutions are known, though a Taylor Series approach gives the answer fairly quickly. I wouldn't expect your answer to have a trivial solution if this 'simpler problem' doesn't have one. – Brevan Ellefsen Mar 17 '17 at 19:37
  • @MichaelRozenberg I will probably try to write it up tomorrow (this has simply been a busy week for me... no time for MSE). Before I do so, is there anything I should know about the limits you are imposing about "doing this by hand"?? Also, am I allowed to reference existing proofs about $a+b=1 \implies a^{2b}+b^{2a}\le1$ to conclude the result? – Brevan Ellefsen Mar 17 '17 at 19:40
  • @Brevan Ellefsen I think I have a proof of your inequality. – Michael Rozenberg Mar 17 '17 at 19:44
  • @MichaelRozenberg Thanks! I'm glad you liked my attempt, and wish I could've come up with a more analytical and rigorous solution, especially something that doesn't require a calculator. – Robin Aldabanx Mar 18 '17 at 00:32
  • 1
    @Robin Aldabanx You claim "we see that there does not exist an $x_n$ s.t. $n>3$ and $g'(x_n)=0$". Isn't that just like saying "we plotted g and see that $g \leq 1$"? I don't think that this is a valid proof. – cafaxo Mar 26 '17 at 21:54
  • @BrevanEllefsen You said $a+b=1 \implies a^{2b}+b^{2a} \le 1$ is pretty known in the literature; just curious, is there some inequality literature? Or in the Analysis literature (as far as I know, Analysis is the field which deals most with inequalities, but please correct me if I'm wrong). – Ovi Apr 14 '18 at 03:01
  • @Ovi hmmm. Interesting question. Honestly, I have no idea where I first saw the result, it just comes up often enough I considered it a more common inequality. While I could point you towards some beginning references, inequalities are not my field of study (I focus more on where complex analysis and operator theory meet). Honestly, I'd ask the the OP Michael Rosenberg. He's probably asked and answered more inequality questions on this site than anyone, and could point you in the right direction far better than I. – Brevan Ellefsen Apr 14 '18 at 14:43
  • @Ovi your question does remind me that I never finished my answer here. My attempt had some serious flaws at the start, which were actually a result of one of my references having a faulty proof of a lemma I needed! I still think the method of attack I was trying was a lot cleaner than the more numeric approaches currently posted. Perhaps I'll have to revisit this post and give it more thought. – Brevan Ellefsen Apr 14 '18 at 14:46

Let $f \colon [0,1] \rightarrow \mathbb{R}$ be the function given by $$f(x) = x^{4 (1-x)^2}.$$ We will show that if $x \in [0,\frac{1}{2}]$ then $f(x) + f(1-x) \leq 1$.

If $x=0$ then it is clear that $f(0)+f(1)=1$.

Claim 1. If $x \in (0, \frac{71}{200}]$ then $f(x) + f(1-x) \leq 1$.

Proof. Let $x \in (0, \frac{71}{200}]$. By Bernoulli's inequality we have $$f(1-x) = (1-x)^{4x^2} \leq 1-4x^3.$$ Therefore \begin{align} &f(x) + f(1-x) \leq 1 \\ &\impliedby f(x) \leq 4x^3 \\ &\iff \log x^{4 (1-x)^2} \leq \log4x^3 \\ &\iff (3-4 (1-x)^2) \log x\ + \log 4 \geq 0. \end{align} The $\log x$ is a problem here. We use the following trick to factor our expression.

Since the global maximum of $z \mapsto - z \log z$ is $\mathrm{e}^{-1}$, we have $(-\mathrm{e}\, x \log x) \log 4 \leq \log 4.$ It follows that \begin{align} &(3-4 (1-x)^2) \log x\ + \log 4 \geq 0 \\ &\impliedby (3-4 (1-x)^2 -(\mathrm{e}\ \log 4)\, x) \log x\ \geq 0 \\ &\iff 3-4 (1-x)^2 -(\mathrm{e}\ \log 4)\, x \leq 0 \\ &\impliedby x \leq \frac{71}{200} < \frac{1}{4} \left(4-\mathrm{e} \log 2-\sqrt{12+\mathrm{e}^2 \log ^2 2-8 \mathrm{e} \log 2}\right). \end{align} We used the quadratic formula in the last step. $$\tag*{$\Box$}$$

Claim 2. If $x \in (\frac{71}{200}, \frac{73}{200}]$ then $f(x) + f(1-x) < 1$.

Proof. Let $z \in (0,1)$. Since $\log(z) \leq z-1$, we have

\begin{align} f'(z) &= x^{4 (1-z)^2} \left(\frac{4 (1-z)^2}{z}-8 (1-z) \log z\right) \\ &\geq x^{4 (1-z)^2} \left(\frac{4 (1-z)^2}{z}+8 (1-z)^2\right) \geq 0. \end{align}

Thus $f$ is monotonically increasing on $(0,1)$. We have $$f(x) + f(1-x) \leq f\left(\frac{73}{200}\right) + f\left(1-\frac{71}{200}\right) \approx 0.9985 < 1$$ for all $x \in (\frac{71}{200}, \frac{73}{200}]$. $$\tag*{$\Box$}$$

We need some lemmata for $(\frac{73}{200}, \frac{1}{2}]$.

Let $\varphi \colon (-\frac{1}{2},\frac{1}{2}) \rightarrow \mathbb{R}$ be the function given by $\varphi(z) = \log \left(\frac{1}{2}-z\right).$

Lemma 3.1. If $z \in (-\frac{1}{2},\frac{1}{2})$ then $\varphi(z) \leq -\log 2 -2 z-2 z^2-\frac{8}{3}z^3$.

Proof. Let $z \in (-\frac{1}{2},\frac{1}{2})$. Since $\log$ is real analytic, we have \begin{align} \varphi(z) &= \sum_{k=0}^{\infty} \frac{\varphi^{(k)}(0)}{k!}z^k \\&=-\log 2 -2 z-2 z^2-\frac{8}{3}z^3 + \sum_{k=4}^{\infty} \frac{\varphi^{(k)}(0)}{k!}z^k. \end{align} By the Lagrange form of the remainder there is a $\zeta \in (-\frac{1}{2},\frac{1}{2})$ such that $$\frac{\varphi^{(4)}(\zeta)}{4!}z^4 = \sum_{k=4}^{\infty} \frac{\varphi^{(k)}(0)}{k!}z^k.$$ We have \begin{align} \varphi(z) &= -\log 2 -2 z-2 z^2-\frac{8}{3}z^3 + \frac{\varphi^{(4)}(\zeta)}{4!}z^4 \\ &= -\log 2 -2 z-2 z^2-\frac{8}{3}z^3 -\frac{1}{4\left(\frac{1}{2}-\zeta\right)^4} z^4 \\ &\leq -\log 2 -2 z-2 z^2-\frac{8}{3}z^3. \end{align} $$\tag*{$\Box$}$$

Let $\gamma \colon (-\frac{1}{2},\frac{1}{2}) \rightarrow \mathbb{R}$ be the function given by $$\gamma(z) = 4 \left(z+\frac{1}{2}\right)^2 \left(-\log 2 -2 z-2 z^2-\frac{8}{3}z^3\right).$$

Lemma 3.2. If $z \in [-\frac{27}{200},\frac{27}{200}]$ then $$-\frac{1}{2}\left(\frac{\gamma(z)-\gamma(-z)}{2}\right)^2-\log 2\geq \frac{\gamma(z)+\gamma(-z)}{2}.$$ Proof. The inequality is equivalent to $$-128 z^8-448 z^6+z^4 (-440-96 \log 2)+z^2 (-42-168 \log2)+18-18 \log ^2 2 -9 \log 2 \geq 0.$$

We substitute $u = z^2$ and use the quartic formula to calculate the roots of the polynomial. The real roots are near $-0.136139$ and $0.136139$. We have $$-0.136 < -\frac{27}{200} < \frac{27}{200} < 0.136.$$ We calculate that the inequality holds at $z=0$, thus it must hold for all $z \in [-\frac{27}{200},\frac{27}{200}]$. $$\tag*{$\Box$}$$

Let $\psi \colon \mathbb{R} \rightarrow \mathbb{R}$ be the function given by $$\psi(z) = \exp\left(-\frac{z^2}{2}\right)(\exp(z) + \exp(-z)).$$

Lemma 3.3. If $z \in \mathbb{R}$ then $\psi(z) \leq 2$.

Proof. For all $z \in [0, \infty)$ we have \begin{align} \exp\left(\frac{z^2}{2}+z\right)\psi'(z) &= -1 -z -(z-1)\exp(2 z) \\ &= -1 -z -(z-1) \sum_{k=0}^{\infty}\frac{2^k z^k}{k!} \\ &= -1 -z - (z-1) \sum_{k=0}^{\infty}\frac{2^k z^k}{k!} \\ &= -1 -z + \sum_{k=0}^{\infty}\frac{2^k z^k}{k!} - \sum_{k=1}^{\infty}\frac{2^{k-1} z^{k}}{(k-1)!} \\ &= -z + \sum_{k=1}^{\infty}2^{k-1}\left(\frac{2}{k!} - \frac{1}{(k-1)!}\right) z^k \\ &= -z + \sum_{k=1}^{\infty}2^{k-1}\left(\frac{2 (k-1)!-k!}{(k-1)! k!}\right) z^k \\ &= \sum_{k=3}^{\infty}2^{k-1}\left(\frac{2 (k-1)!-k!}{(k-1)! k!}\right) z^k. \end{align} Since $2 (k-1)! < k!$ for all $k > 2$, we have $\psi'(z) \leq 0$. Thus $\psi$ is monotonically decreasing on $[0,\infty)$. We have $\psi(0) = 2$, thus $\psi(z) \leq 2$ for all $z \in [0,\infty)$. Since $\psi(z) = \psi(-z)$ for all $z \in \mathbb{R}$, we have $\psi(z) \leq 2$. $$\tag*{$\Box$}$$

Claim 3.4. If $x \in (\frac{73}{200}, \frac{1}{2}]$ then $f(x) + f(1-x) \leq 1$.

Proof. Let $x \in (\frac{73}{200}, \frac{1}{2}]$ and $z = \frac{1}{2} - x \in [0, \frac{27}{200})$. We have

\begin{align} f(x) + f(1-x) &= f\left(\frac{1}{2}-z\right) +f\left(\frac{1}{2}+z\right) \\[10pt] &= \exp \left(4 \left(\frac{1}{2}+z\right)^2 \varphi(z)\right) + \exp \left(4 \left(\frac{1}{2}-z\right)^2 \varphi(-z)\right) \\ \text{By Lemma 3.1:} \\[7pt] &\leq \exp \left(\gamma(z)\right) + \exp \left(\gamma(-z)\right) \\[10pt] &= \exp \frac{\gamma(z)+\gamma(-z)}{2} \left(\exp \frac{\gamma(z)-\gamma(-z)}{2} + \exp \frac{\gamma(-z)-\gamma(z)}{2}\right) \\ \text{By Lemma 3.2:} \\[7pt] &\leq \exp \left(-\frac{1}{2}\left(\frac{\gamma(z)-\gamma(-z)}{2}\right)^2-\log 2\right) \left(\exp \frac{\gamma(z)-\gamma(-z)}{2} + \exp \frac{\gamma(-z)-\gamma(z)}{2}\right) \\[10pt] &= \frac{1}{2}\, \exp \left(-\frac{1}{2}\left(\frac{\gamma(z)-\gamma(-z)}{2}\right)^2\right) \left(\exp \frac{\gamma(z)-\gamma(-z)}{2} + \exp \frac{\gamma(-z)-\gamma(z)}{2}\right) \\[10pt] &= \frac{1}{2}\, \psi\left(\frac{\gamma(z)+\gamma(-z)}{2}\right) \\ \text{By Lemma 3.3:} \\[7pt] &\leq 1. \end{align} $$\tag*{$\Box$}$$

  • 587
  • 3
  • 9
  • 5
    Impressive solution. The use of Lemma 3.3 particularly nice; here is an alternative proof. We need $(1-z)e^{2z}\leq 1+z$, which is clear for $z>1$. For $z\in[0,1]$ we use $$\ln\left(\frac{1+z}{1-z}\right)=2\left(z+\frac{z^3}3+\frac{z^5}5+\cdots\right)\geq2z.$$ – chronondecay Mar 24 '17 at 12:34

I offer a complete, self-contained solution below that can be checked without computer/calculator assistance.

The main tools that we will use (abuse?) are logarithmic differentiation and symmetry, to greatly simplify the exponentials that we are dealing with.

We may assume that $0<a\leq\frac12\leq b<1$. We split into the following two cases.

Case 1: $a\leq\frac13$

By Bernoulli's inequality we have $b^{4a^2}=(1-a)^{4a^2}\leq1-4a^3$, so it suffices to show that $$a^{4(1-a)^2}\stackrel?<4a^3.$$ Let $h(a)=(4(1-a)^2-3)\ln(a)$; we want to show that $h(a)\stackrel?<\ln4$. Now $$h'(a)=\frac{4(1-a)^2-3}a-8(1-a)\ln(a)$$ can clearly be seen to be decreasing on $a\in[0,1]$ (edit: in fact this is false as stated, but writing $\frac{4(1-a)^2-3}a=\frac{(2a-1)^2}a-4$ we see that $h'(a)$ is decreasing for $a\in[0,\frac12]$). Hence $$h'(a)\geq h'(\frac13)=\frac{16\ln3-11}3>\frac{16-11}3>0,$$ so $h(a)$ is increasing on $a\in[0,\frac13]$. Thus $$h(a)\leq h(\frac13)=\frac{11}9\ln3<\ln4,$$ since $4^4=256>243=3^5$ implies $\frac{\ln4}{\ln3}>\frac54>\frac{11}9$, as desired.

Case 2: $\frac13\leq a\leq\frac12$

Substitute $a=\frac12-x$ and $b=\frac12+x$, so $x\in[0,\frac16]$ and $$b^{4a^2}+a^{4b^2}=\left(\frac12+x\right)^{(1-2x)^2}\!\!\!+\left(\frac12-x\right)^{(1+2x)^2}\!\!\!=F(x)+F(-x),$$ where $$F(x)=\left(\frac12+x\right)^{(1-2x)^2}\!\!\!.$$

Write $F'(x)=F(x)G(x)$ (so $G$ is the logarithmic derivative of $F$). It is clear that $F(x)$ is increasing on $x\in[-\frac16,\frac16]$, so $F(x),F'(x)>0$ implies $G(x)>0$ on $x\in[-\frac16,\frac16]$.

Now $F(0)+F(-0)=1$, so \begin{align*} F(x)+F(-x)\stackrel?\leq1 &\:\Longleftarrow\:\frac d{dx}(F(x)+F(-x))\stackrel?\leq0\\ &\iff F'(x)\stackrel?\leq F'(-x)\\ &\iff\frac{F(x)}{F(-x)}\stackrel?\leq\frac{G(-x)}{G(x)}. \end{align*} We now prove this last inequality on $x\in[0,\frac16]$ in the following steps.

Step 1

We will show that $\ln\left(\dfrac{F(x)}{F(-x)}\right)$ is concave up on $x\in[0,\frac16]$, ie. \begin{align*} \frac{d^2}{dx^2}\ln\left(\dfrac{F(x)}{F(-x)}\right)&=\frac d{dx}(G(x)+G(-x))\\ &=G'(x)-G'(-x)\stackrel?\geq0. \end{align*} It suffices to show that $G'(x)$ is increasing on $x\in[-\frac16,\frac16]$. Now \begin{align*} G(x)&=\frac d{dx}\ln(F(x))\\ &=(1-2x)\left(\frac{1-2x}{\frac12+x}-4\ln\left(\frac12+x\right)\right),\\ G'(x)=\cdots&=\frac43\cdot\frac{-4+9(x+\frac16)^2}{(\frac12+x)^2}+8\ln\left(\frac12+x\right),\tag{*} \end{align*} so (by some miracle!) $G'(x)$ can be clearly seen to be increasing on $x\in[-\frac16,\frac16]$.

Hence the graph of $\ln\left(\dfrac{F(x)}{F(-x)}\right)$ lies below the line joining $(0,0)$ and $\left(\dfrac16,\ln\left(\dfrac{F(\frac16)}{F(-\frac16)}\right)\right)$, ie. $$\ln\left(\dfrac{F(x)}{F(-x)}\right)\leq6\ln\left(\dfrac{F(\frac16)}{F(-\frac16)}\right)x=2Cx,\quad C:=\frac{4\ln54}3.$$

Step 2

We are left to show that $\dfrac{G(-x)}{G(x)}\stackrel?\geq e^{2Cx}$ on $x\in[0,\frac16]$.

A little manipulation gives \begin{align*} \frac{G(x)}{\sqrt{1-4x^2}}&=\sqrt{\frac{1-2x}{1+2x}}\left(2\,\frac{1-2x}{1+2x}-4\ln\left(\frac12+x\right)\right)\\ &=u(2u^2+4\ln(1+u^2)):=H(u), \end{align*} under the substitution $u=\sqrt{\dfrac{1-2x}{1+2x}}$, $x=\dfrac12\,\dfrac{1-u^2}{1+u^2}$.

Now the desired inequality is equivalent to $$\frac{H(\frac1u)e^{-Cx}}{H(u)e^{Cx}}=\frac{G(-x)e^{-Cx}}{G(x)e^{Cx}}\stackrel?\geq1\quad\text{for }x\in[0,\tfrac16]\iff u\in[\tfrac1{\sqrt2},1].$$ Note that $u$ is decreasing in $x$, and the transformation $x\to-x$ is equivalent to $u\to\frac1u$. Thus it is sufficient to show that $H(u)e^{Cx}$ is increasing on $u\in[\frac1{\sqrt2},\sqrt2]$ (this is less than clear; I have gotten the sign wrong several times myself), ie. \begin{align*} \frac d{du}\ln(H(u)e^{Cx})&=\frac d{du}(Cx+\ln H(u))\\ &=C\frac{dx}{du}+\frac{H'(u)}{H(u)}\\ &=-2C\frac u{(1+u^2)^2}+\frac1u+\frac{2u+\frac{4u}{1+u^2}}{u^2+2\ln(1+u^2)}\stackrel?\geq0\\ \iff\frac{2C}{1+u^2}&\stackrel?\leq\frac{1+u^2}{u^2}+\frac{2(3+u^2)}{u^2+2\ln(1+u^2)}. \end{align*} We substitute $t=u^2$ and use the inequality $\ln(1+t)\leq\ln2+\frac{t-1}2$ (by convexity of $\ln(t+1)$; RHS is tangent line at $t=1$) to reduce the above to $$2+\frac1t+\frac{7-2\ln2}{2t+2\ln2-1}\stackrel?\geq\frac{2C}{1+t}.\tag{**}$$ This will fall to Cauchy-Schwarz (in the Engel form $\sum\dfrac{a_i^2}{b_i}\geq\dfrac{(\sum a_i)^2}{\sum b_i}$), if we can find the correct weights. With some (okay, a lot) of inspiration, we obtain the following: \begin{align*} &\phantom{{}={}}2+\frac1t+\frac{7-2\ln2}{2t+2\ln2-1}\\ &\geq\frac{94}{47}+\frac{18}{18t}+\frac{101}{36+7t}\\ &\geq\frac{(\sqrt{94}+\sqrt{18}+\sqrt{101})^2}{54(1+t)}\geq\frac{2C}{1+t}, \end{align*} which is what we wanted.

Hence we are left with the following (zero-variable!) inequalities that we used above: $$\ln2\stackrel?\leq\frac{25}{36},\qquad\frac{(\sqrt{94}+\sqrt{18}+\sqrt{101})^2}{54}\stackrel?\geq2C.$$ Most people should be content with checking these by calculator. For the purists, here is a sketch of how to get these by hand.

Step 3 (Optional?)

Firstly, note that for $x>0$, we have \begin{align*} \ln\left(\frac{1+x}{1-x}\right)&=2\left(x+\frac{x^3}3+\frac{x^5}5+\frac{x^7}7+\cdots\right)\\ &\leq2\left(x+\frac{x^3}3+\frac{x^5}3+\frac{x^7}3+\cdots\right)\\ &=2\left(x+\frac{x^3}{3(1-x^2)}\right). \end{align*} Taking $x=\frac13$ gives the first inequality. In addition, taking $x=\frac15$ gives $$54(2C)=54\times\frac83\left(3\ln\frac32+\ln2\right)\leq72\left(3\times\frac{73}{180}+\frac{25}{36}\right)=\frac{2876}5.$$ Let $(a,b,c,d)=(101,94,18,\frac{2876}5)$, and $n=\frac12(c+d-a-b)$. We are left to check that \begin{align*} \sqrt a+\sqrt b+\sqrt c&\geq\sqrt d\\ \iff a+b+2\sqrt{ab}&\geq c+d-2\sqrt{cd}\qquad(\because\sqrt a+\sqrt b>0)\\ \iff\sqrt{ab}&\geq n-\sqrt{cd}\\ \iff ab&\geq n^2+cd-2n\sqrt{cd}\qquad(\because\sqrt{ab}>0)\\ \iff2n\sqrt{cd}&\geq n^2+cd-ab\\ \iff4n^2cd&\geq(n^2+cd-ab)^2,\qquad(\because2n\sqrt{cd}>0) \end{align*} which after substitution becomes $\dfrac{205212545208}{125}\geq\dfrac{16402832101681}{10000}$, something that I consider to be close to the limits of what I can do by hand.

QED. Phew!


  1. Yes, this might be a long proof, and perhaps it doesn't give much insight as to why the original inequality holds. However, I hope that the tricks used in the solution (and there are many!) are of independent interest.

  2. I have optimised the solution as presented above to minimise the effort of hand calculations. Some parts of the solution have feasible alternatives, eg. at $(*)$ we can compute $G''(x)=\dfrac{16(7+8x+4x^2)}{(1+2x)^3}$, and check that the numerator is positive on $x\in[-\frac16,\frac16]$; at $(**)$ we essentially need $P(t)\geq0$ for some cubic $P$, and we can proceed by showing $$P(t)\geq k(t-\lambda)(t-\mu)^2+Q(t)$$ for good choices of $k,\lambda,\mu\in\mathbb R$ and $Q$ a quadratic in $t$.

  3. How human is this solution? It is certainly human-checkable, but I have serious doubts as to whether a human can come up with a proof along these lines without computer assistance. As many previous answers have noted, the original inequality is very tight, and we cannot afford to lose more than $0.03$ in total throughout our proof.

    In addition, I can count about 8 places in Step 2 alone where I used the irreversible $\Leftarrow$ implication, ie. "to prove A it is sufficient to prove B." This is disastrous if B turns out to be false! I personally needed extensive computer aid to explore the problem space and avoid dead ends of this type --- this proof was made possible by around 2000--3000 Mathematica commands.

    For these two reasons, I doubt that there can be an unassisted proof using the normal tools of calculus (such as those in the above solution), let alone in a competition setting. Of course, maybe we just need some radical new insight or perspective. (I have not looked at the $W$ function closely, so maybe...?)


  • 423
  • 2
  • 10
  • 1
    Thank you for your trying! – Michael Rozenberg Mar 20 '17 at 04:37
  • 1
    In Case 1, you claim $h'$ is clearly decreasing for $a\in(0,1]$. First of all, this is false. A plot of $h''$ reveals positive output for $a$ just below $1$, so $h'$ is increasing a little bit toward the end of $(0,1]$. You might respond by restricting the claim to $(0,1/3]$. But then it's still not "clear" than $h'$ is decreasing. Trying to prove something like that is the crux of this whole endeavor. It's trying to prove an inequality involving a hybrid polynomial logarithmic function. – 2'5 9'2 Mar 20 '17 at 19:27
  • @alex.jordan Thanks for catching that, I have edited the solution to include an easy fix. – chronondecay Mar 21 '17 at 01:10


I want to share some thought. Consider the more general problem $$ a^{n b^2} + b^{n a^2} \leq 1 $$ A key observation is the symmetry of the two terms $a^{n b^2} $ and $b^{n \ a^2} $. Due to the constraint $a + b = 1 $, the two terms are just $a^{n (1 - a)^2} $ and $ (1 - a)^{n a^2} $. So a substitution of $a \rightarrow 1 - a$ changes one term to the other. Conclusion : The LHS is a sum of two terms symmetric around the $a = 1/2 $.

Lemma W.l.o.g, suppose a function $f (x) $ in interval $[0, 1] $.If the function is monotonic and convex around $x=1/2$, then the "mirror average function" $g (x) = (f (x) + f (1 - x))/2 $ has a maximum at $x = 1/2 $.

Proof Just calculate to show $g' (1/2) = 0 $ and $g'' (x) = f'' (x)$

Corollary for concave $f(x)$ follows directly. This analysis does not answer the question, but hopefully will introduce some abstractness and shed more light on it.

Old Post

This should be a comment, but then I won't be able to post pictures. For the more general inequality:

$$ a^{n b^2}+b^{n a^2}\leq1 $$

I draw pictures for n = 0, 1, ..., 7. Each one has green dashed contour highlighting where the equality is satisfied. And of course, each one is overlaid with $a+b=1$. It's interesting to note only $n = 4$ is tightly bounded by the green contour, so it is a really special $n$ value.

integer n from 0 to 7, inclusive

  • 2,561
  • 10
  • 17

This is indeed a hard nut, since convexity cannot be invoked to close the case. The following plot shows that the function $$f(x):=(1-x)^{4x^2}+x^{4(1-x)^2}\qquad(0\leq x\leq 1)$$ in fact never drops below $0.97\>$! (Compare Robin Aldabanx' answer)

enter image description here

At the moment I'm just able to show that $f(x)$ behaves as claimed near $x=0$ (and, by symmetry, near $x=1$) and near $x={1\over2}$.

If $0\leq x\leq{1\over2}$ then $0\leq4x^2\leq1$, and Bernoulli's inequality gives $$(1-x)^{4x^2}\leq1-4x^3\ .$$ On the other hand $$x^{4(1-x)^2}=x^4\cdot x^{-8x+4x^2}=:x^4\> h(x)$$ with $\lim_{x\to0+}h(x)=1$. It follows that there is a $\delta>0$ with $$f(x)\leq 1-4x^3+2x^4=1-4x^3\>\left(1-{x\over2}\right)<1\qquad (0<x<\delta)\ .$$ For $x\doteq{1\over2}$ we consider the auxiliary function $$g(t):=f\left({1\over2}+t\right)\qquad\bigl(|t|\ll1\bigr)$$ which is analytic for small $|t|$. Mathematica computes its Taylor series as $$g(t)=1+\left(-8+4\log 2+8\log^2 2\right) t^2+\ ?\>t^3\ .$$ The numerical value of the relevant coefficient here is $\doteq-1.38379$, and this tells us that $f$ has a local maximum at $x={1\over2}$.

Christian Blatter
  • 216,873
  • 13
  • 166
  • 425
  • 4
    Your first estimate works for $0 < x \le \frac 14$ (unless I made an error), because $x^{4(1-x)^2} \le x^{-8x} x^4 \le 16 x^4$ on that interval, so that $f(x) \le 1 - 4x^3 + 16 x^4 = 1 - 4x^3(1-4x) \le 1$. – Martin R Mar 17 '17 at 20:29
  • 1
    "This is a hard nut ..." as a teenager, I did not view this maturely. – Mr Pie Jun 04 '20 at 03:49

Too long for a comment .

It's just to propose one methods on power series .

Method (Power series)

First we have two inequalities ($\forall x\in[\frac{42}{100},\frac{1}{2}]$)

$$(1-x)^{4x^2}\leq \frac{1}{2}+\frac{2}{5} (\frac{1}{2}x-x^2)+ (\frac{-(2+4\ln(\frac{1}{2}))}{2}+\frac{18}{10}) (\frac{1}{2}-x)\quad(1)$$


$$x^{4(1-x)^2}\leq \frac{1}{2}+\frac{2}{5} (\frac{1}{2}x-\frac{1}{4})+ (\frac{-(2+4\ln(\frac{1}{2}))}{2}+\frac{18}{10}+\frac{2}{5}(x-\frac{1}{2}))(x-\frac{1}{2})$$

proof of $(1)$ :

Taking logartihm on both side then making the difference and differentiate and using :

$$8x\ln(1-x)\geq (0.5-\frac{2235}{1000}(0.5-x))\ln(0.5)$$

We get a fourth degree polynomial with a roots at $x=0.5$.Remains to study a third degree polynomial wich is not hard .

The proof of $(2)$ is similar .

For the case $\forall x\in[0,\frac{1}{3}]$ we can use the Bernoulli's inequality as in others answer . Remains just to prove the inequality on $[\frac{1}{3},\frac{45}{100}]$.I continue to explore this .

Another approach :

We use a form of the Young's inequality wich is a somewhere a generalization of the Bernoulli's inequality :

Let $a,b>0$ and $0<v<1$ then we have :

$$av+b(1-v)\geq a^vb^{1-v}$$

Taking account of this theorem and putting :

$a=x^{2(1-x)}$$\quad$$b=1$$\quad$$v=2(1-x)$ we get $0.5\leq x<1$:

$$x^{4(1-x)^2}\leq x^{2(1-x)}2(1-x)+1-2(1-x)$$

Now the idea is to show :

Let $$(1-x)^{4x^2}\leq 1-(x^{2(1-x)}2(1-x)+1-2(1-x))$$

Or :


Or: $$(1-x)^{4x^2-1}+2x^{2(1-x)}\leq 2$$

Now by Bernoulli's inequality we have :

$$2x^{2(1-x)}\leq 2-4(1-x)^2$$

So :

$$(1-x)^{4x^2-1}+2-4(1-x)^2\leq 2$$

Or :

$$(1-x)^{4x^2-3}\leq 4$$

It's not hard to show it on $[0.65,1)$

As you can see here we can improve the reasoning above and get the inequality on $[0.61,1]$

Update 20/04/2021

We can improve the reasoning using the inequality :

$$(1-v)a+vb\geq a^{1-v}b^{v}+\min{(1-v,v)}(\sqrt{a}-\sqrt{b})^2$$

Found in this article (Proposition 1.1) https://arxiv.org/pdf/1001.0535.pdf

Edit :

We have the inequality for $0<x\leq 0.55$

$$\left(\frac{1}{1-\left(1-\frac{1}{1-\left(1-x\right)^{2\left(x\right)}}\right)\left(2^{-y}x^{-y}\right)}\right)\geq x^{4\left(1-x\right)^{2}}\quad (I)$$

Where $y=\frac{-(2+4\ln(\frac{1}{2}))}{2}+1$

We have also $0<x<1$:

$$r(x)=\left(\left(\frac{1}{1+\left(\frac{g\left(x\right)}{1-\left(1-x\right)^{2x}}-g\left(x\right)\right)2^{-f\left(x\right)}x^{-f\left(x\right)}}\right)^{-1}-1+g\left(x\right)\right)^{-1}\geq x^{4\left(1-x\right)^{2}}$$

Where :


And :


To finish we have for $\frac{1}{4}\leq x\leq \frac{3}{4}$:

$$r(x)+r(1-x)\leq 1$$

A good start to show the inequality $(I)$ see WA.The same idea can be applied to the next inequality.

To simplify a little bit $(I)$ we have the inequality for $0.47\leq x\leq 0.53$ :

$$\left(1+\left(\frac{1}{1-\left(2^{\left(2\left(1-x\right)+1\right)}\cdot\left(1-x\right)^{2}\cdot x\right)}-1\right)\cdot2^{-\left(-\frac{\left(2+4\ln\left(0.5\right)\right)}{2}+1\right)}\cdot x^{-\left(-\frac{\left(2+4\ln\left(0.5\right)\right)}{2}+1\right)}\right)\cdot x^{4\left(1-x\right)^{2}}\leq 1$$

Last edit :

It seems we have on $x\in[0.45,0.55]$ :

$$\frac{\left(2^{-\left(2\left(1-x\right)\right)^{2}}\cdot x\right)}{1-2^{\ln\left(2\right)}\left(\left(1-x\right)\cdot2\cdot x\right)^{\left(\ln\left(2\right)+1\right)}}\geq x^{\left(2\left(1-x\right)\right)^{2}}$$

Erik Satie
  • 3,402
  • 2
  • 6
  • 29

Again too long for comment

The inequality is equivalent to on $(0,0.5)$:


Now define :


Where $m(x)=x^{-\left(2\left(1-x\right)\right)^{2}}$ and $g\left(x\right)=\left(\left(1-x\right)^{-\left(2x\right)^{2}}-1\right)$

We have the inequalities :

$$f(x)+(u(x)-1)g(x)\geq 2$$

$$f(x)\geq \frac{\left(\left(u\left(1-x\right)\right)\cdot\left(u\left(x\right)\right)\right)^{0.25}}{\sqrt{2}}\geq 1$$ This kind of method is valid for a generalisation when the higher exponent (here $2$ in $f(x)$) is greater or equal to $2$.

Erik Satie
  • 3,402
  • 2
  • 6
  • 29