$a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$

Question

$a, b,c $ are positive real numbers such that $a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$

Any ideas ?

Answer 1

We need to prove that $$\sum_{cyc}a^2\sqrt{a^2+3}\geq6$$ for non-negatives $a$, $b$ and $c$ such that $a^2+b^2+c^2=3$.

Indeed, by C-S $$\sum_{cyc}a^2\sqrt{a^2+3}=\frac{1}{2}\sum_{cyc}a^2\sqrt{(1+3) (a^2+3)}\geq\frac{1}{2}\sum_{cyc}a^2(a+3)=\frac{1}{2}(a^3+b^3+c^3)+\frac{9}{2}.$$ Thus, it's enough to prove that $$a^3+b^3+c^3\geq3,$$ which is Power-Means: $$\left(\frac{a^3+b^3+c^3}{3}\right)^2\geq\left(\frac{a^2+b^2+c^2}{3}\right)^3.$$

Done!

Answer 2

By Holder $$\left(\sum_{cyc}a\sqrt{a+3}\right)^2\sum_{cyc}\frac{a}{a+3}\geq(a+b+c)^3=27.$$ Thus, it remains to prove that $$\sum_{cyc}\frac{a}{a+3}\leq\frac{3}{4}$$ or $$\sum_{cyc}\left(\frac{1}{4}-\frac{a}{a+3}\right)\geq0$$ or $$\sum_{cyc}\frac{1-a}{a+3}\geq0$$ or

$$\sum_{cyc}\left(\frac{1-a}{a+3}+\frac{1}{4}(a-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(a-1)^2}{a+3}\geq0.$$ Done!

Answer 3

That's Jensen's inequality for the convex function $x\,\sqrt{x+3}=(x+3)^{3/2}-3\,\sqrt{x+3}$.