a+b+c=3 inequality

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a+b+c=3 inequality

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#1 h Mar 27, 2025, 5:41 AM

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Let a,b,c\ge 0: ab+bc+ca>0 then prove \frac{b+c}{a^{2}+bc}+\frac{c+a}{b^{2}+ca}+\frac{a+b}{c^{2}+ab}+\frac{a+b+c}{ab+bc+ca}\ge \frac{12}{a+b+c}.}

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#2 h Mar 27, 2025, 8:37 AM

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JK1603JK wrote:

Let $a,b,c\ge 0$ : $ab+bc+ca>0$ then prove $\frac{b+c}{a^{2}+bc}+\frac{c+a}{b^{2}+ca}+\frac{a+b}{c^{2}+ab}+\frac{a+b+c}{ab+bc+ca}\ge \frac{12}{a+b+c}.$

According to the C-S inequality,
$\sum\left(\frac{(b+c)(a+b+c)}{a^2+bc} + 1\right)=\sum (a^2+bc) \sum \frac{1}{a^2+bc} + 2\sum \frac{bc}{a^2+bc} \geq 9+2\frac{(\sum bc)^2}{\sum (a^2+bc)bc}.$ Thus, it suffices to prove
$\frac{2(\sum bc)^2}{\sum (a^2+bc)bc}+\frac{(a+b+c)^2}{ab+bc+ca}\geq 6.$ Moreover,
$2(\sum bc)^3+(a+b+c)^2\Bigl(\sum (a^2+bc)bc\Bigr)-6\Bigl(\sum (a^2+bc)bc\Bigr)(\sum bc)=\sum a^2b^2(a-b)^2+abc\sum a(a-b)(a-c)\geq 0.$ Therefore, the original inequality holds.

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