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Today I'm interested by the following problem :

Let $x,y>0$ then we have : $$x+y-\sqrt{xy}\leq\exp\Big(\frac{x\ln(x)+y\ln(y)}{x+y}\Big)$$

The equality case comes when $x=y$

My proof uses derivative because for $x\geq y $ the function : $$f(x)=x+y-\sqrt{xy}-\exp\Big(\frac{x\ln(x)+y\ln(y)}{x+y}\Big)$$

is decreasing and for $y\geq x$ the function is increasing and the maximum occurs when $x=y$

My question is : Have you an alternative proof wich doesn't use derivative ?

Thanks in advance.

Alex Ravsky
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  • I rewrote the RHS as $\sqrt[x+y]{x^x y^y}$ and showed that this was greater than or equal to $\frac{x+y}{2}$ (briefly: apply weighted AM-GM to $1/x$ and $1/y$ with respective weights $x$ and $y$, and then invert). But this doesn't help, as it reduces proving the original inequality to showing that $\frac{x+y}{2} \leq \sqrt{xy}$, which obviously isn't true. – Connor Harris Jun 06 '19 at 13:52
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    This is not new, actually (https://math.stackexchange.com/questions/1432043/an-upper-bound-of-binary-entropy/1432228#1432228) – Jack D'Aurizio Jul 29 '19 at 21:39
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    @JackD'Aurizio: The two inequalities do not look the same, though. – Hans Nov 05 '19 at 23:30
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    This problem is equivalent to showing that given Riesz conjugates $p$ and $q$ (i.e. $1/p+1/q=1$) then the inequality below holds $$\sqrt{pq}+p^{1/p}q^{1/q}\geq p^{1/2+1/p}q^{1/2+1/q}$$ Maybe someone can get Holder's Inequality to make it work. –  Nov 13 '19 at 03:47
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    @JackD'Aurizio : https://math.stackexchange.com/questions/1432043/an-upper-bound-of-binary-entropy/1432228#1432228 is too weak – user90369 Nov 14 '19 at 10:35
  • I did it! Waiting for comments. – Yuri Negometyanov May 21 '21 at 16:06

6 Answers6

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This is not an answer to the question, but it's too big to put it to the comment. I will show some connection (which might be interesting) between this inequality and Shannon entropy.

Firstly rewrite this expression as $$ x + y- \sqrt{xy} \le x^{\frac{x}{x+y}} \cdot y^{\frac{y}{x+y}}. $$ Then one can use a substitution $$ a = \frac{x}{x+y}, \; b = \frac{y}{x+y} $$ and get an equivalelent inequality in terms of $a$ and $b$ $$ (1 - \sqrt{ab}) \le a^a b^b, \;\; a +b = 1. $$ So, we need to show that given $a+b = 1$, we will have $$ \sqrt{ab} + a^a b^b \ge 1. $$ It's equivalent to the following upper bound for Shannon entropy $H(a,b)$ $$ H(a,b) = -a \log a - b\log b \le -\log(1-\sqrt{ab}), \;\; a+b=1. $$

So, one needs to prove this estimate for Shannon entropy $H(a,b)$. Unfortunately, I have no idea how to do this without calculus. Plots of $H(a,b)$ and its upper bound:

enter image description here

It might happen that this inequality has some meaning in information theory, though I haven't found anything about that.

Virtuoz
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Incomplete answer

This is a trick that sometimes works when dealing with inequalities with two variables; however, in this case, the prohibition of calculus makes the problem more difficult.

Let $y=ax$ for some $a,x>0$. Then \begin{align}x+y-\sqrt{xy}\leq\exp\left(\frac{x\ln x+y\ln y}{x+y}\right)&\impliedby x+ax-x\sqrt a\le\exp\left(\frac{x\ln x+ax\ln ax}{x+ax}\right)\\&\impliedby x(1-\sqrt a+a)\le\exp\left(\ln x+\frac{a\ln a}{1+a}\right)\\&\impliedby1-\sqrt a+a\le a^{\frac a{1+a}}\end{align} so it suffices to show that $(1-\sqrt a+a)^{a+1}\le a^a$ for all $a\in(0,1)$, where $y<x$ without loss of generality.

It may be worth noting that the inequality is extremely tight which can be seen via this visualisation.

TheSimpliFire
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  • Since we are working with positive values it might be conceptually easier to substitute $a=b^2$. I tried expanding the exponential, which is acceptable since it's one of the many definitions of exp, but no finite number of terms seems to give the bound for all $x$, which suggests that any method not involving differentiation is likely to fail. – Morgan Rogers Jun 13 '19 at 07:47
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A COMMENT.

Actualy holds the following (somewhat) conjecture generalization

If $N>1$ and $x_1,x_2,\ldots,x_N>0$, then $$ \frac{1}{N}\sum^{N}_{k=1}x_k-\sqrt[N]{\prod^{N}_{k=1}x_k}\leq\exp\left(\frac{\sum^{N}_{k=1}x_k\log x_k}{\sum^{N}_{k=1}x_k}\right)\tag 1 $$

Nikos Bagis
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Idea for a proof :

As the inequality is homogenous we obtain an equivalent inequality like :

$$\left(\frac{1}{x}\right)^{\frac{2}{x^2+1}}+\frac{1}{x}-\frac{1}{x^2}\geq1$$

Using the Bernoulli's first approximation and some others terms in the Newton's expansion of $(1+x)^a$ we have ($0< x\leq 1$).:

$$\left(\frac{1}{x}\right)^{\frac{2}{x^{2}+1}}+\frac{1}{x}-\frac{1}{x^{2}}\geq 1+\frac{1}{6}\cdot\left(\frac{2}{x^{2}+1}-2\right)\left(\frac{2}{x^{2}+1}-1\right)\left(\frac{2}{x^{2}+1}\right)\left(\frac{1}{x}-1\right)^{3}+\left(\frac{1}{x}-1\right)\left(\frac{2}{x^{2}+1}\right)+\frac{1}{x}-\frac{1}{x^{2}}+0.5\left(\frac{2}{x^{2}+1}-1\right)\left(\frac{2}{x^{2}+1}\right)\left(\frac{1}{x}-1\right)^{2}\geq 1$$

The RHS becomes (using Wolfram alpha for the simpplification) :

$$ \frac{1}{3}\frac{(3x^6-14x^5+27x^4-28x^3+17x^2-6x+1)}{(x(x^2+1)^3)}\geq 0$$

Or :

$$\frac{1}{3}\frac{(x-1)^4 (3x^2-2x+1)}{(x(x^2+1)^3)}\geq 0$$

Wich is obvious !

Done !

For a reference of proof without calculus of the binomial theorem see :

Reference :

https://www.jstor.org/stable/2319010?seq=1

Binomial theorem proof for rational index without calculus

Erik Satie
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This is not an answer but a remark that I think everyone on this topic has missed so far.( That is why I posted as an answer)

The ‘derivative’ solution presented by OP’s teacher is WRONG. The function $f(x)$ is not decreasing when $x\le y$ . More precisely, $x=0$ should be another maxima of this function as $$\lim_{x\rightarrow 0^+}f(x)=0$$

That is also why the ‘solutions’ given by @Yuri and @erik deserve more recognition.

Paresseux Nguyen
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The alternative forms of the given inequality are $$x\ln x + y \ln y \ge (x+y)\ln(x+y)+(x+y)\ln\left(1-\dfrac{\sqrt{xy}}{x+y}\right),$$ $$x\ln\,\dfrac x{x+y} + y \ln\, \dfrac y{x+y} \ge (x+y)\ln\left(1-\dfrac{\sqrt{xy}}{x+y}\right),$$ $$\dfrac x{x+y}\,\ln\,\dfrac x{x+y} + \dfrac y{x+y}\, \ln\, \dfrac y{x+y} \ge \ln\left(1-\sqrt{\dfrac x{x+y}}\;\sqrt{\dfrac y{x+y}}\right).\tag1$$ Let WLOG $\;x\le y,\;$ $$\dfrac yx = \dfrac{1+z}{1-z},\quad \dfrac x{x+y} =\dfrac{1-z}{2} ,\quad \dfrac y{x+y} =\dfrac{1+z}{2},\quad z\in[0,1],\tag2$$ then $(1)$ can be presented in the forms of $$\dfrac{1-z}{2}\,\ln\dfrac{1-z}{2}+\dfrac{1+z}{2}\,\ln\dfrac{1+z}{2} \ge \ln(1-\,^1\!/_2\sqrt{1-z^2}\,),$$ $$\dfrac{1-z}{2}\,\ln(1-z)+\dfrac{1+z}{2}\,\ln(1+z) \ge \ln(2-\sqrt{1-z^2}),$$ $$-\dfrac{z}{2}\,\ln(1-z)+\dfrac{z}{2}\,\ln(1+z) \ge \ln(2-\sqrt{1-z^2}) -\ln\sqrt{1-z^2},$$ $$f(z) = (1-z)^{-\large\,^z\!/_2}(1+z)^{\large\,^z\!/_2}+1- \dfrac 2{\sqrt{1-z^2}}\ge 0. \tag3$$ Using binomial series in the forms of $$(1-z)^{-\large\,^z\!/_2} = 1 + \dfrac12 z^2 + \dfrac1{2!}\,\dfrac z2\,\dfrac{2+z}2 z^2 + \dfrac1{3!}\,\dfrac z2\,\dfrac{2+z}2\dfrac{4+z}2 z^3 +\dots$$ $$= 1+\dfrac12z^2+\dfrac14z^3+\dfrac7{24}z^4+\dfrac14z^5+\dfrac{113}{480}z^6+\dots,$$ $$(1+z)^{\large\,^z\!/_2} = 1 + \dfrac12 z^2 + \dfrac1{2!}\,\dfrac z2\,\dfrac{-2+z}2 z^2 + \dfrac1{3!}\,\dfrac z2\,\dfrac{-2+z}2\dfrac{-4+z}2 z^3 +\dots$$ $$= 1+\dfrac12z^2-\dfrac14z^3+\dfrac7{24}z^4-\dfrac14z^5+\dfrac{113}{480}z^6+\dots,$$ $$(1-z^2)^{-\large\,^1\!/_2} = 1+\dfrac12z^2+\dfrac38z^4+\dfrac5{16}z^6+\dots,$$ one can get the Maclaurin series in the form of $$f(z) = \left(1+\dfrac12z^2+\dfrac7{24}z^4+\dfrac{113}{480}z^6+\dots\right)^2 -\left(\dfrac14z^3+\dfrac14z^5+\dots\right)^2$$ $$+1-2\left(1+\dfrac12z^2+\dfrac38z^4+\dfrac5{16}z^6+\dots\right),$$ $$f(z) = \dfrac1{12}z^4+\dfrac3{40}z^6+\dots$$ (see also WA series),

with the all positive coefficients.

Proved!

$\color{brown}{\textbf{About binomial series.}}$

  • Newton's binomial formulas exist independently of the derivatives associated theory.

  • If the exponent is the rational, then the coeffitients of the binomial series $(1+x)^{\large\frac pq} = 1+a_1x+a_2x^2+\dots$ are defined via the identity $$(1+a_1x+a_2x^2+\dots)^q = (1+x)^q,$$ by the comparation of the polynomials in the LHS and RHS.

  • If the exponent is real, then the binomial series are defined via the limiting transition.

These elementary knowledges point that the binomial series exist independently of the derivatives associated theory.

Yuri Negometyanov
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  • Using a Maclaurin Series is just a more complicated way of using derivatives. The OP already knows you can show that using calculus on the derivative you can get the result. – Eric May 22 '21 at 03:42
  • @Eric You are wrong. However, thanks for the comment - note about binomial series added. – Yuri Negometyanov May 22 '21 at 05:03