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

Solution 1:

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.

Solution 2:

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.

Solution 3:

We will begin with an observation. Consider the function $$f(a,b,c)=\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} .$$ It has the property that $$f(ta,tb,tc)=f(a,b,c).$$ We may thus assume without loss of generality that $$a+b+c=1$$, by setting $$t=\frac{1}{a+b+c}.$$ Under this assumption, we may rewrite the function as $$\sqrt{\frac{2a}{1-a}}+\sqrt{\frac{2b}{1-b}}+\sqrt{\frac{2c}{1-c}}$$ If $a=1-\epsilon$ for $\epsilon>0$, $b+c=\epsilon$. Now note that in order for these square roots to be defined over the reals, then $0\leq a,b,c<1$, which means that $b,c<\epsilon$. Now for such an $\epsilon$, $$\sqrt{\frac{2b}{1-b}}, \sqrt{\frac{2c}{1-c}}$$ are both close to zero, and $$\sqrt{\frac{2a}{1-a}}$$ is large.