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pennypc123456789

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pennypc123456789

#1 h Apr 29, 2025, 2:42 PM

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Given $a,b,c$ be positive real numbers . Prove that
$\frac{ab}{(a+b)^2} +\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2} \ge \frac{6abc }{(a+b)(b+c)(a+c)}$

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#2 h Apr 29, 2025, 3:35 PM

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pennypc123456789 wrote:

Given $a,b,c$ be positive real numbers . Prove that
$\frac{ab}{(a+b)^2} +\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2} \ge \frac{6abc }{(a+b)(b+c)(a+c)}$

Let $t = \frac{abc}{(a+b)(b+c)(a+c)}$ . We have $t\leq \frac{1}{8}$ . Using Am-Gm inequality, we get:
$\frac{ab}{(a+b)^2} +\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2} \geq 3\sqrt[3]{t^2}\geq 6t = \frac{6abc }{(a+b)(b+c)(a+c)}$ Equality at $a=b=c$ .

This post has been edited 1 time. Last edited by GeoMorocco, Apr 30, 2025, 6:55 AM

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pennypc123456789

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pennypc123456789

#3 h Apr 29, 2025, 11:59 PM

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GeoMorocco wrote:

pennypc123456789 wrote:

Given $a,b,c$ be positive real numbers . Prove that
$\frac{ab}{(a+b)^2} +\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2} \ge \frac{6abc }{(a+b)(b+c)(a+c)}$

Let $t = \frac{abc}{(a+b)(b+c)(a+c)}$ . We have $t\leq \frac{1}{8}$ . Using Am-Gm inequality, we get:
$\frac{ab}{(a+b)^2} +\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2} \geq 3\sqrt[3]{t}\geq 6t = \frac{6abc }{(a+b)(b+c)(a+c)}$ Equality at $a=b=c$ .

How to prove $3\sqrt[3]{t}\geq 6t$ ?

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#4 h Apr 30, 2025, 6:57 AM • 1 Y

Y by pennypc123456789

there was a typo, it should be $3\sqrt[3]{t^2}\geq 6t$ , which can be proved as follow:
$3\sqrt[3]{t^2}\geq 3\sqrt[3]{t^2.8t} = 6t$

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