An interesting inequality

$\large \frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{1}{3}\geq\frac{8}{9}\bigg(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\bigg)$ Let $a,b,c>0$ , prove the above inequality.

This problem was taken from a discussion on VMF (Viet Maths Forum)

#Algebra

Easy Math Editor

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Comments

Here's the solution. First, without losing generality, we can assume that $ab+bc+ac=3$ , the problem'll become $a^2+b^2+c^2+1\geq\frac{8}{3}\bigg(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg)$ Now we'll do some manipulation on the variables in $LHS$ $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a(a+b)(a+c)+b(b+a)(b+c)+c(c+a)(c+b)}{(a+b)(b+c)(c+a)}$ $=\frac{a^3+b^3+c^3+(a+b+c)(ab+bc+ca)}{(a+b+c)(ab+bc+ca)-abc}=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca+ab+bc+ca)+3abc}{3(a+b+c)-abc}$ $=\frac{(a+b+c)(a^2+b^2+c^2)+3abc}{3(a+b+c)-abc}$ So now the inequality is $a^2+b^2+c^2+1\geq\frac{8[(a+b+c)(a^2+b^2+c^2)+3abc]}{9(a+b+c)-3abc}$ $\Rightarrow [9(a+b+c)-3abc](a^2+b^2+c^2+1)\geq 8(a+b+c)(a^2+b^2+c^2)+24abc$ $\Leftrightarrow 9(a+b+c)(a^2+b^2+c^2)+9(a+b+c)-3abc(a^2+b^2+c^2)-3abc\geq 8(a+b+c)(a^2+b^2+c^2)+24abc$ $\Leftrightarrow (a+b+c)(a^2+b^2+c^2+9)\geq 3abc(a^2+b^2+c^2+9)$ $\Rightarrow a+b+c\geq 3abc$ The last inequality can be proven easily by these following inequalities: $(a+b+c)^2\geq 3(ab+bc+ca)$ and $abc\leq\sqrt{\big(\frac{ab+bc+ca}{3}\big)^3}$ . The equality holds when $a=b=c$

P C - 5 years ago

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Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
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`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$