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How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$ for $a,b,c>0$ and $abc=1$?

I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$

Indeed,$\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$

$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$ $\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$ $\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!) It is wrong. Advice on solving this problem.

Michael Rozenberg
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Other
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  • We can at least part of the way by using $AM \geq GM$, $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge 3 \sqrt[3]{\frac{abc}{(1+a)(1+b)(1+c)}} = 3 \frac{1}{\sqrt[3]{(1+a)(1+b)(1+c)}}$$ Not sure if this is a useful stepping stone. – Simon S Nov 01 '14 at 09:55
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    @SimonS - how is that true? the root should be cubic, not square. – nbubis Nov 01 '14 at 09:57
  • My idea is a stepping stone in the wrong direction: the *upper* bound on $1/\sqrt[3]{(1+a)(1+b)(1+c)}$ is $\sqrt{2}/2$ when that's what we want the *lower* bound to be. – Simon S Nov 01 '14 at 10:42

7 Answers7

3

Let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$, where $x$, $y$ and $z$ are positives.

Hence, we need to prove that $$\sum_{cyc}\frac{\frac{x}{y}}{\sqrt{1+\frac{x}{y}}}\geq\frac{3}{\sqrt2}$$ or $$\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\geq\frac{3}{\sqrt2}.$$ Now, by Holder $$\left(\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\right)^2\sum_{cyc}xy(x+y)(2z+x+y)^3\geq\left(\sum_{cyc}x(2z+x+y)\right)^3.$$ Thus, it remains to prove that $$2\left(\sum_{cyc}(x^2+3xy)\right)^3\geq9\sum_{cyc}xy(x+y)(2z+x+y)^3$$ or $$\sum_{sym}(x^6+9x^5y+24x^4y^2+18x^3y^3+9x^4yz-36x^3y^2z-25x^2y^2z^2)\geq0,$$ which is obviously true.

Done!

Michael Rozenberg
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2

Consider the function for positive $x$: $$f(x) = \frac{x}{\sqrt{1+x}}-\frac1{\sqrt 2}-\frac3{4\sqrt 2}\log x$$

Note that $f(x) \ge 0 \implies f(a)+f(b)+f(c) \ge 0 \implies $ the given inequality. Now $$f'(x) = \frac{4x^2-3\sqrt2 (x+1)^{3/2}+8x}{8x(x+1)^{3/2}}$$

We need to check the sign of the numerator, $4(x+1)^2-3\sqrt2(x+1)^{3/2}-4$. Using $y = \sqrt{x+1}$, we get the numerator as

$$4y^4-3\sqrt2y^3-4 = (y-\sqrt2)(4y^3+\sqrt2y^2+2y+2\sqrt2)$$

As the second factor is positive, the numerator's sign is given by $y-\sqrt2$ which has the same sign as $x-1$, so $f'(x)< 0$ for $x < 1$ and $f'(x)> 0$ for $x> 1$. Hence $f(x)\ge f(1)=0$.

Macavity
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By the same way as Mr. Mike$,$ it's enough to prove $$2\left(\sum_{cyc}(x^2+3yz)\right)^3\geqslant 9\sum xy(x+y)(2z+x+y)^3$$ Or $$\frac18\sum \left( 16\,{x}^{4}+100\,{z}^{4}+104\,{x}^{3}y+243\,{y}^{2}{z}^{2 }+330\,{z}^{3}x+416\,{y}^{2}zx+342\,{z}^{2}xy \right) \left( x-y \right) ^{2}+$$ $$+\frac18\sum x{y}^{2} \left( 18\,y+41\,x \right) \left( z+x-2\,y \right) ^{2}\geqslant 0$$

tthnew
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By Hölder's inequality $$\left(\frac a{\sqrt{1+a}}+\frac b{\sqrt{1+b}}+\frac c{\sqrt{1+c}}\right)^2\Big(a(1+a)+b(1+b)+c(1+c)\Big)\ge(a+b+c)^3,$$ so we only need to prove \begin{align*}(a+b+c)^3&\ge\frac92\Big(a(1+a)+b(1+b)+c(1+c)\Big)\\(a+b+c)(2(a+b+c)^2-9)&\ge9(a^2+b^2+c^2)\\\end{align*} AM-GM tells us that $(a+b+c)^2=(a+b+c)^2+9-9\ge6(a+b+c)-9$, so it suffices to prove \begin{align*}(a+b+c)(12(a+b+c)^2-27)&\ge9(a^2+b^2+c^2)\\4(a+b+c)^2&\ge3(a^2+b^2+c^2)+9(a+b+c)\\a^2+b^2+c^2+8(ab+bc+ca)&\ge9(a+b+c)\end{align*} And I'll leave this last inequality up to you. Use the fact that $abc=1$, which haven't been used so far.

user2345215
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Let$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}$$ $$g(a,b,c)=abc-1=0$$

Using Lagrange Multiplier $$\large\frac{\frac{\partial f}{\partial a}}{\frac{\partial g}{\partial a}}= \frac{\frac{\partial f}{\partial b}}{\frac{\partial g}{\partial b}}= \frac{\frac{\partial f}{\partial c}}{\frac{\partial g}{\partial c}}=k $$ We get $$\frac{a+2}{2bc(a+1)^{3/2}}=\frac{b+2}{2ac(b+1)^{3/2}}=\frac{c+2}{2ab(c+1)^{3/2}}=k$$

by solving this for $a,b,c$ we get $$a=b=c$$ and from constraint $g(a,b,c)=0$ we get $$a=b=c=1$$ $$f_{min}=\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{1+1}}=\frac{3}{\sqrt2}=\frac{3\sqrt{2}}{2}$$ $$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}\ge f_{min}$$

$$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}\ge \frac{3\sqrt{2}}{2}$$

0

Thus, it remains to prove that $$2\left(\sum_{cyc}(x^2+3xy)\right)^3\geq9\sum_{cyc}xy(x+y)(2z+x+y)^3$$

We have \begin{align*}LHS-RHS&=(x^3+y^3+z^3)\sum x(x-y)(x-z)+(x^2+y^2+z^2-xy-yz-zx)^3\\ &+7(xy+yz+zx)\sum x^2(x-y)(x-z)\\ &+[x^2+y^2+z^2+3(xy+yz+zx)][x^2y^2+y^2z^2+z^2x^2-xyz(x+y+z)]\\ &+ (x+y+z)[4(x^2+y^2+z^2)+9(xy+yz+zx)]\sum z(x-y)^2\\ &+ (x+y+z)(xy+yz+zx)\sum (x+y+7z)(x-y)^2 \ge 0\end{align*}

BestChoice123
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Some observations:

$$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \tag{1}$$

$$A=\sqrt{1+a}+\sqrt{1+b}+\sqrt{1+c} +\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} -\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)$$ $$A\ge(2+2+2)-\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)\tag{2}$$

Comparing with the original definition: $$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2},\tag{3}$$

We conclude that we need to prove:

$$\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} \le 3-\frac{3\sqrt{2}}{4}\tag{4}$$

mike
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