confusing inequality

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#1 h Apr 18, 2025, 2:07 PM

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Let $a,b,c>0$ such that: $a^2b^2+ c^2b^2+ a^2c^2=3(abc)^2$ . Prove that: $\sum \frac{b+c}{a} \geq 2\sqrt{3(ab+bc+ca)}$

This post has been edited 3 times. Last edited by giangtruong13, Apr 20, 2025, 3:02 PM

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#2 h Apr 18, 2025, 4:14 PM

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giangtruong13 wrote:

Let $a,b,c>0$ such that: $a^2b^2+ c^2b^2+ a^2c^2=3(abc)^2$ . Prove that: $\sum \frac{a}{b+c} \geq 2\sqrt{3(ab+bc+ca)}$

It's $\sum_{cyc}\frac{a}{b+c}\geq2\sqrt{\frac{(ab+ac+bc)(a^2b^2+a^2c^2+b^2c^2)}{a^2b^2c^2}}.$ Are you sure that it's true?

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#3 h Apr 19, 2025, 3:03 PM

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Oh sorry, i write wrongly, i will fix it here :oops_sign:

:oops_sign:

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#4 h Apr 20, 2025, 2:41 AM

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Bummppppp

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#5 h Apr 20, 2025, 2:57 PM

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This inequality was from a book by an inactive user $toanmuonmau$

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#6 h Apr 20, 2025, 7:57 PM • 1 Y

Y by kiyoras_2001

giangtruong13 wrote:

Let $a,b,c>0$ such that: $a^2b^2+ c^2b^2+ a^2c^2=3(abc)^2$ . Prove that: $\sum \frac{b+c}{a} \geq 2\sqrt{3(ab+bc+ca)}$

Because by C-S we obtain:
$\sum_{cyc}\frac{b+c}{a}=\frac{\sum\limits_{cyc}(a^2b+a^2c)}{abc}=\frac{\sum\limits_{cyc}a^2\sum\limits_{cyc}a-\sum\limits_{cyc}a^3}{abc}=$ $=\frac{\sqrt{\left(\sum\limits_{cyc}a^2\right)^2\left(\sum\limits_{cyc}a\right)^2}-\sum\limits_{cyc}a^3}{abc}=\frac{\sqrt{\sum\limits_{cyc}(a^4+2a^2b^2)\sum\limits_{cyc}(a^2+2ab)}-\sum\limits_{cyc}a^3}{abc}\geq$ $\geq\frac{\sqrt{\sum\limits_{cyc}a^4\sum\limits_{cyc}a^2}+2\sqrt{\sum\limits_{cyc}a^2b^2\sum\limits_{cyc}ab}-\sum\limits_{cyc}a^3}{abc}\geq\frac{2\sqrt{\sum\limits_{cyc}a^2b^2\sum\limits_{cyc}ab}}{abc}=2\sqrt{3(ab+ac+bc)}.$

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