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#1 h Apr 1, 2025, 9:54 AM • 2 Y

Y by arqady, cubres

Let $a$ , $b$ , $c$ be non-negative real numbers, no two of which are zero. Prove that :
$\sum \frac{3ab-2bc+3ca}{3b^2+bc+3c^2} \geq \frac{12}{7}$

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arqady

#2 h Apr 1, 2025, 1:45 PM • 2 Y

Y by SunnyEvan, cubres

SunnyEvan wrote:

Let $a$ , $b$ , $c$ be non-negative real numbers, no two of which are zero. Prove that :
$\sum \frac{3ab-2bc+3ca}{3b^2+bc+3c^2} \geq \frac{12}{7}$

Let $a\geq b\geq c$ . Thus,
$\sum_{cyc}\frac{3ab-2bc+3ca}{3b^2+bc+3c^2} -\frac{12}{7}=\sum_{cyc}\left(\frac{3ab-2bc+3ca}{3b^2+bc+3c^2} -\frac{4}{7}\right)=$ $=\frac{3}{7}\sum_{cyc}\frac{7ab+7ac-6bc-4b^2-4c^2}{3b^2+bc+3c^2} =\frac{3}{7}\sum_{cyc}\frac{(a-b)(4b+3c)-(c-a)(4c+3b)}{3b^2+bc+3c^2} =$ $=\frac{3}{7}\sum_{cyc}(a-b)\left(\frac{4b+3c}{3b^2+bc+3c^2}-\frac{4a+3c}{3c^2+ac+3a^2}\right)=$ $=\frac{3\sum\limits_{cyc}(a-b)^2(3a^2+ab+3b^2)(12ab+9ac+9bc-9c^2)}{7\prod\limits_{cyc}(3a^2+ab+3b^2)}\geq\frac{27\sum\limits_{cyc}(a-b)^2(3a^2+ab+3b^2)c(a+b-c)}{7\prod\limits_{cyc}(3a^2+ab+3b^2)}\geq$ $\geq\frac{27\left((a-c)^2(3a^2+ac+3c^2)b(a-b)+(b-c)^2(3b^2+bc+3c^2)a(b-a)\right)}{7\prod\limits_{cyc}(3a^2+ab+3b^2)}\geq$ $\geq\frac{27\left((b-c)^2(3a^2+ac+3c^2)b(a-b)+(b-c)^2(3b^2+bc+3c^2)a(b-a)\right)}{7\prod\limits_{cyc}(3a^2+ab+3b^2)}=$ $=\frac{81(b-c)^2(a-b)^2(ab-c^2)}{7\prod\limits_{cyc}(3a^2+ab+3b^2)}\geq0.$

This post has been edited 1 time. Last edited by arqady, Apr 1, 2025, 1:46 PM

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#3 h Apr 1, 2025, 1:47 PM • 1 Y

Y by cubres

Nice solution! :-D

:-D

This post has been edited 1 time. Last edited by SunnyEvan, Apr 7, 2025, 12:41 PM

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#4 h Apr 3, 2025, 8:55 AM • 2 Y

Y by SunnyEvan, cubres

equivalent to
$7\sum{ab(a-b)^2(3a^2+ab+3b^2)}+6\prod{(a-b)^2}\ge0$

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#9 h Apr 12, 2025, 8:57 AM • 1 Y

Y by cubres

DKI wrote:

equivalent to
$7\sum{ab(a-b)^2(3a^2+ab+3b^2)}+6\prod{(a-b)^2}\ge0$

Thanks for your help.

This post has been edited 1 time. Last edited by SunnyEvan, Apr 12, 2025, 9:48 AM

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#10 h Apr 12, 2025, 9:11 AM • 1 Y

Y by cubres

Let $a$ , $b$ , $c$ be non-negative real numbers, no two of which are zero. Prove that :
$\sum \frac{-ab+3bc-ca}{2b^2+3bc+2c^2} \leq \frac{3}{7}$

This post has been edited 2 times. Last edited by SunnyEvan, Apr 12, 2025, 11:29 AM

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#13 h Apr 12, 2025, 2:57 PM • 1 Y

Y by cubres

HELP

:help:

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