Inequality

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#1 h Jan 22, 2021, 5:51 AM

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Let $a,b,c>0$ .Prove that $\dfrac{1}{(a+b)\sqrt{(a+2c)(b+2c)}}>\dfrac{3}{2(a+b+c)^2}$

This post has been edited 1 time. Last edited by math2000, Jan 22, 2021, 6:12 AM
Reason: Error

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#2 h Jan 24, 2021, 3:33 AM • 1 Y

Y by Euler01

Here's a fairly straightforward attempt using AM-GM

Restructure a bit:
$\begin{align*} \dfrac{1}{(a+b)\sqrt{(a+2c)(b+2c)}}>\dfrac{3}{2(a+b+c)^2} \\[2em] 3(a+b) \sqrt{(a+2c)(b+2c)}<2(a+b+c)^2 \end{align*}$ By AM-GM:

$\sqrt{(a+2c)(b+2c)} \leqslant \frac{a+b+4c}{2}$

Therefore,
$\begin{align*} 3(a+b) \sqrt{(a+2c)(b+2c)} &\leqslant 3(a+b) \left ( \dfrac{a+b+4c}{2} \right ) < 2(a+b+c)^2 \\ \end{align*}$ We let $u = a+b$ :
$\begin{align*} 3u(u+4c) & <4(u+c)^2 \\ 3u^2+12uc & <4u^2+8uc+4c^2 \\ 0 & <(u-2c)^2 \\ 0 & < (a + b - 2c)^2 \end{align*}$

This post has been edited 1 time. Last edited by simorno, Jan 24, 2021, 3:38 AM

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Quantum_fluctuations

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Quantum_fluctuations

#3 h Jan 24, 2021, 5:04 AM • 1 Y

Y by Mango247

Using A.M.-G.M. inequality , we have

$\frac{(3a+3b) +(3a+3b)+(2a+4c)+(2b+4c)}{4} \enspace \geq \enspace \sqrt[4]{(3a+3b)(3a+3b)(2a+4c)(2b+4c)}$

$\iff \frac{8a+8b+8c}{4} \enspace \geq \enspace \sqrt{6(a+b)} \enspace \cdot \enspace \sqrt[4]{ (a+2c)(b+2c)}$
$\iff 4(a+b+c)^2 \enspace \geq \enspace 6(a+b)\sqrt{ (a+2c)(b+2c)}$
$\iff \frac{1}{ (a+b)\sqrt{ (a+2c)(b+2c)}} \enspace \geq \enspace \frac{3}{2(a+b+c)^2}$
which is just the desired inequality.

This post has been edited 1 time. Last edited by Quantum_fluctuations, Jan 24, 2021, 5:05 AM

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sqing

#4 h Jan 24, 2021, 6:25 AM

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math2000 wrote:

Let $a,b,c>0$ .Prove that $\dfrac{1}{(a+b)\sqrt{(a+2c)(b+2c)}}>\dfrac{3}{2(a+b+c)^2}$

Let $a,b,c>0$ .Prove that $(a+b+c)^2 \enspace \geq \enspace \frac 32(a+b)\sqrt{ (a+2c)(b+2c)}$

This post has been edited 1 time. Last edited by sqing, Jan 24, 2021, 6:27 AM

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#5 h Jan 24, 2021, 9:08 AM • 1 Y

Y by simorno

simorno wrote:

$0 < (a + b - 2c)^2$

There may be a case of equality here. For equality, we must have: $a+b=2c$ , putting this into initial inequality:
$\dfrac{1}{2c\sqrt{(a+2c)(b+2c)}}>\dfrac{3}{2(2c+c)^2}$ After simple transformation:
$9c^2 = (a+2c)(b+2c)$ $9c^2 = ab + 2c(a+b) + 4c^2$ Since $a+b =c$ , we have:
$c^2 = ab$ Given that $2c = a+b$ , we get that equality is achieved when $a=b=c$ , which indeed fits.

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#6 h Jan 27, 2021, 11:46 PM • 1 Y

Y by Mango247

sqing wrote:

math2000 wrote:

Let $a,b,c>0$ .Prove that $\dfrac{1}{(a+b)\sqrt{(a+2c)(b+2c)}}>\dfrac{3}{2(a+b+c)^2}$

Let $a,b,c>0$ .Prove that $(a+b+c)^2 \enspace \geq \enspace \frac 32(a+b)\sqrt{ (a+2c)(b+2c)}$

By AM-GM, we have $\sqrt{(a+2c)(b+2c)}\leq \frac{a+b}{2}+2c$ . Therefore, we have
$\begin{align*} &(a+b+c)^2\geq \frac{3}{2}(a+b)\sqrt{(a+2c)(b+2c)} \\ \iff &(a+b+c)^2 \geq \frac{3}{2}(a+b)(\frac{a+b}{2}+2c) \\ \iff &(a+b)^2+4c^2\geq 2\cdot 2c (a+b) \end{align*}$ which is just AM-GM again, and the equality holds when $a:b:c=1:1:2$ . $\blacksquare$

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sqing

#7 h Apr 10, 2025, 4:05 AM

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https://artofproblemsolving.com/community/c6h2139038p34505795

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imnotgoodatmathsorry

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imnotgoodatmathsorry

#8 h Apr 12, 2025, 2:59 AM

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Keith50 wrote:

sqing wrote:

math2000 wrote:

Let $a,b,c>0$ .Prove that $\dfrac{1}{(a+b)\sqrt{(a+2c)(b+2c)}}>\dfrac{3}{2(a+b+c)^2}$

Let $a,b,c>0$ .Prove that $(a+b+c)^2 \enspace \geq \enspace \frac 32(a+b)\sqrt{ (a+2c)(b+2c)}$

By AM-GM, we have $\sqrt{(a+2c)(b+2c)}\leq \frac{a+b}{2}+2c$ . Therefore, we have
$\begin{align*} &(a+b+c)^2\geq \frac{3}{2}(a+b)\sqrt{(a+2c)(b+2c)} \\ \iff &(a+b+c)^2 \geq \frac{3}{2}(a+b)(\frac{a+b}{2}+2c) \\ \iff &(a+b)^2+4c^2\geq 2\cdot 2c (a+b) \end{align*}$ which is just AM-GM again, and the equality holds when $a:b:c=1:1:2$ . $\blacksquare$

$a:b:c=1:1:1$ *

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