Inequality

by utkarshgupta, Feb 28, 2015, 2:20 PM

Problem :
Let a, b, c be positive real numbers such that $ ab + bc + ac \ge 3 $. Prove that

$ \frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{a+c}} \ge \frac{3}{\sqrt{2}} $.

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Solution :


Let $ S=\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{a+c}} $
By Holder's Inequality,
\[S^2(a(a+b)+b(b+c)+c(c+a)) \ge (a+b+c)^3\]
\[\implies S^2(a^2+b^2+c^2+ab+bc+ca) \ge (a+b+c)^3\]
\[ \implies S^2 \ge \frac{(a+b+c)^3}{a^2+b^2+c^2+ab+bc+ca}\]

Setting $u=a+b+c$, $v=ab+bc+ca \ge 3$
\[\implies S^2 \ge \frac{u^3}{u^2-v}\]

But by AM-GM, $2u^3+27 \ge 9u^2$
$\implies 2u^3 + 9v \ge 9u^2$
$\implies \frac{u^3}{u^2-v} \ge \frac{9}{2}$

\[\implies S^2 \ge \frac{9}{2}\]
\[\implies S\ge \frac{3}{\sqrt{2}}\]

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