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

Denote $$S = \sqrt{\frac{a}{b+c}} + \sqrt{\frac{b}{c+a}} + \sqrt{\frac{c}{a+b}}$$

The Inequality doesn't hold. Clearly, taking $b=c=0.05, a=5$ implies $$\sqrt{2}S>\sqrt{\frac{2a}{b+c}} = 10 > 3$$ By this method, it is easily shown that no upper bound exists.

For a lower bound, note firstly that when $a = b, c \rightarrow 0$, $S \rightarrow 2$. We'll prove that $S \ge 2$ for non negative reals $a,b,c$.

We have, $$\frac{a+b+c}{2a} = \frac 12 \left(\frac{b+c}{a} + 1 \right) \ge \sqrt{\frac{b+c}{a}} \\ \implies \sqrt{\frac{a}{b+c}} \ge \frac{2a}{a+b+c}$$ Adding up the other two similar inequalities, we get the result.


You got it wrong. It should be: $$ \sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}} \leq 3 $$

See my comment on Inequality. $\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3$ for a proof.

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