Prove an inequality with square roots

Question

Prove that $$\frac{a}{\sqrt{b+c}} + \frac{b}{\sqrt{c+a}} + \frac{c}{\sqrt{a+b}} > \sqrt{a+b+c}$$ if a, b, c are positive.

Answer 1

$$\frac{a}{\sqrt{b+c}} > \frac{a}{\sqrt{a+b+c}} $$

since $a+b+c > b+c$. Similarly

$$\frac{b}{\sqrt{c+a}} > \frac{b}{\sqrt{a+b+c}} $$ $$\frac{c}{\sqrt{a+b}} > \frac{c}{\sqrt{a+b+c}} $$

Combine these to get: $$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} > \frac{a+b+c}{\sqrt{a+b+c}} = \sqrt{a+b+c}$$

Answer 2

You can proceed as in the proof of the similar inequality

• $\frac{a}{\sqrt{a+b}} + \frac{b}{\sqrt{b+c}} + \frac{c}{\sqrt{c+a}} > \sqrt{a+b+c}$ is true for positive a,b,c

$$ \frac{a}{\sqrt{b+c}} + \frac{b}{\sqrt{c+a}} + \frac{c}{\sqrt{a+b}} \\> \frac{a}{\sqrt{a+b+c}} + \frac{b}{\sqrt{b+c+a}} + \frac{c}{\sqrt{c+a+b}} \\= \frac{a+b+c}{\sqrt{a+b+c}} = \sqrt{a+b+c} $$

Answer 3

Hint: Set $a+b+c=3$, then RHS $=\sqrt3$. Then use the convex function $\dfrac{x}{\sqrt{3-x}}$ with Jensen to prove a tighter inequality, $LHS\geqslant \sqrt{\frac32}RHS$.