If, $a+b+c+abc=4$, with $a,b,c$ being positive reals, then prove or disprove the following inequality: $$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}}+\frac{c}{\sqrt{a+b}}\geq\frac{a+b+c}{\sqrt2}$$

I couldn't do anything, please help. However, from the some values I have tried, the inequality does seem to be true.


We have by Holder's inequality: $$\left(\sum_{cyc} \frac{a}{\sqrt{b+c}}\right)^2\left(\sum_{cyc}a(b+c) \right) \ge (a+b+c)^3$$

So it is sufficient to show that $$2(a+b+c) \ge \sum_{cyc} a(b+c) \text{ or equivalently, } a+b+c \ge ab + bc + ca$$

Suppose $a+b+c < ab + bc + ca$. Then by Schur's inequality we have $$\begin{align} \frac{9abc}{a+b+c} &\ge 4(ab+bc+ca)- (a+b+c)^2 \\ &> (a+b+c)\left( 4- (a+b+c)\right) \\ &= (a+b+c) \cdot abc \\ \end{align}$$

This gives $a+b+c< 3$, further we have from $4= a+b+c+ abc \ge 4\sqrt{abc} \implies abc \le 1$, so $4 = a+b+c+abc < 3+1$, a contradiction. Hence proved.

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    $\begingroup$ You should find references on the web, here is one (artofproblemsolving.com/Wiki/index.php/…). In this problem we can avoid Holder by doing Cauchy-Schwarz twice, but Holder seems more natural. $\endgroup$ – Macavity Apr 20 '14 at 3:14

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