Inequality about non-negative variables

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Inequality about non-negative variables

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#1 h Mar 27, 2025, 12:25 PM

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Let a,b,c\ge 0: ab+bc+ca>0 then prove \frac{(a+b)(a+c)}{b^2+c^2}+\frac{(b+c)(b+a)}{c^2+a^2}+\frac{(c+a)(c+b)}{a^2+b^2}\ge \frac{18(ab+bc+ca)}{(a+b+c)^2}

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SomeonecoolLovesMaths

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#2 h Mar 27, 2025, 12:54 PM

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JK1603JK wrote:

Let $a,b,c\ge 0 \: ab+bc+ca>0$ then prove $\frac{(a+b)(a+c)}{b^2+c^2}+\frac{(b+c)(b+a)}{c^2+a^2}+\frac{(c+a)(c+b)}{a^2+b^2}\ge \frac{18(ab+bc+ca)}{(a+b+c)^2}$

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mathlover_forever

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#3 h Mar 27, 2025, 1:39 PM

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than kyou

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#4 h Mar 28, 2025, 1:32 AM

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Can someone solve it?

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#5 h Mar 28, 2025, 3:31 AM

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Let $u = a^2 + b^2 + c^2$ and $v = ab + bc + ca$ .

Since
$\frac{(a+b)(a+c)}{b^2+c^2} + 1 = \frac{u+v}{b^2+c^2},$ the original inequality is equivalent to
$(u+v) \sum \frac{1}{b^2+c^2} \ge \frac{18v}{u+2v} + 3 = \frac{3(u+8v)}{u+2v}.$
Dividing both sides by $u+v$ , we obtain
$\sum \frac{1}{b^2+c^2} \ge \frac{3(u+8v)}{(u+2v)(u+v)}.$
Multiplying both sides by $u$ , we have
$u\sum \frac{1}{b^2+c^2} \ge \frac{3(u+8v)u}{(u+2v)(u+v)}.$
Since
$\frac{u}{b^2+c^2} = 1 + \frac{a^2}{b^2+c^2},$ this is equivalent to
$\sum \frac{a^2}{b^2+c^2} \ge \frac{3(u+8v)u}{(u+2v)(u+v)} - 3 = \frac{3(5u-2v)v}{(u+2v)(u+v)}.$
By the Cauchy–Schwarz inequality,
$\sum \frac{a^2}{b^2+c^2} \ge \frac{(\sum a^2)^2}{\sum a^2(b^2+c^2)} = \frac{u^2}{2\sum a^2b^2}.$
Notice that
$7u^2 - 5uv + 6v^2 - 24\sum a^2b^2 = 7\sum a^2(a-b)(a-c) + 2\sum ab(a-b)^2 \ge 0.$
Hence,
$2\sum a^2b^2 \le \frac{7u^2-5uv+6v^2}{12},$ so that
$\frac{u^2}{2\sum a^2b^2} \ge \frac{12u^2}{7u^2-5uv+6v^2}.$
It suffices to prove that
$\frac{12u^2}{7u^2-5uv+6v^2} \ge \frac{3(5u-2v)v}{(u+2v)(u+v)},$ which is equivalent to
$3(4u-3v)(u-2v)^2(u-v) \ge 0.$ This inequality is evidently true.

Therefore, the original inequality holds.

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