13
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Simon S Nov 1 '14 at 9:55
  • 1
    $\begingroup$ @SimonS - how is that true? the root should be cubic, not square. $\endgroup$ – nbubis Nov 1 '14 at 9:57
  • $\begingroup$ 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. $\endgroup$ – Simon S Nov 1 '14 at 10:42
2
$\begingroup$

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 $f(x)\ge f(1)=0$.

$\endgroup$
2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ obviously true ? what about the $-36$ and $-25$ coefficients ? $\endgroup$ – Ewan Delanoy Apr 16 '17 at 17:01
  • $\begingroup$ @Ewan Delanoy Since $(6,0,0)\succ(3,2,1)$, $(5,1,0)\succ(3,2,1)$, $(4,2,0)\succ(3,2,1)$, $(3,3,0)\succ(3,2,1)$, $(3,3,0)\succ(2,2,2)$ and $(4,1,1)\succ(2,2,2)$, it's just Muirhead. $\endgroup$ – Michael Rozenberg Apr 16 '17 at 17:08
  • $\begingroup$ Got it, thanks for your explanation $\endgroup$ – Ewan Delanoy Apr 16 '17 at 17:19
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $$a=b=c=1$$ $$f_{min}=\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{1+1}}+\frac{1}{\sqrt{1+1}}=\frac{3}{\sqrt2}=\frac{3\sqrt{2}}{2}$$ $$f(a,b,c)=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}}\ge f_{min}$$

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

$\endgroup$
  • $\begingroup$ Why does $ ... = ... = ... = k$ imply that $a = b = c$? $\endgroup$ – Simon S Nov 1 '14 at 14:04
  • $\begingroup$ ...Except if the minimum is reached on the boundary. $\endgroup$ – Did Nov 1 '14 at 14:12
-1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.