An inequality

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An inequality

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#1 h Mar 24, 2025, 5:07 AM

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Let a,b,c\ge 0: ab+bc+ca=1. Prove that \frac{a}{\left(a+1\right)^{2}}+\frac{b}{\left(b+1\right)^{2}}+\frac{c}{\left(c+1\right)^{2}}\ge \frac{1}{a+b+c}+\frac{abc}{4}.

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#2 h Mar 24, 2025, 5:46 AM

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JK1603JK wrote:

Let a,b,c\ge 0: ab+bc+ca=1. Prove that \frac{a}{\left(a+1\right)^{2}}+\frac{b}{\left(b+1\right)^{2}}+\frac{c}{\left(c+1\right)^{2}}\ge \frac{1}{a+b+c}+\frac{abc}{4}.

Let $a,b,c \geq 0: ab+bc+ca=1.$ Prove that
$\frac{a}{(a+1)^2} + \frac{b}{(b+1)^2} + \frac{c}{(c+1)^2} \geq \frac{1}{a+b+c} + \frac{abc}{4}.$

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#3 h Mar 24, 2025, 7:03 AM

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Let $p = a + b + c$ , $q = ab + bc + ca = 1$ , $r = abc$ .
If $p \leq 2$ , then by C-S

$\sum \frac{a}{(a+1)^2} \geq \frac{1}{2} \sum \frac{a}{a^2 + 1} = \frac{1}{2} \sum \frac{a}{(a+b)(a+c)} = \frac{q}{pq - r} = \frac{1}{p - r}.$
Thus, it suffices to show that

$\frac{1}{p - r} \geq \frac{1}{p} + \frac{r}{4},$
which is equivalent to

$p r^2 + r(4 - p^2) \geq 0,$
which obviously holds.

If $p > 2$ , multiplying both sides by $p + 2$ , since $ap \leq ap+bc = a^2 + 1$ , we have

$\sum \frac{a(p + 2)}{(a + 1)^2}=\sum \frac{ap + 2a}{a^2 + 1 + 2a} \geq \sum \frac{ap}{a^2 + 1} = \frac{2p}{p - r}.$
Thus, it suffices to show that

$\frac{2p}{p - r} \geq \frac{p + 2}{p} + \frac{r(p + 2)}{4},$
which is equivalent to,

$(p^2 - 4)\left( (r - 1)^2 + 1 - rp \right) + 2(p + 2)r^2 + 2(p - 2)^2 \geq 0,$
which holds because $1 - rp = q^2 - rp \geq 2rp \geq 0$ .

This post has been edited 1 time. Last edited by lbh_qys, Mar 24, 2025, 7:05 AM

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