Inequality with radicals

Let $a, b, c$ be positive real numbers such that $a^2+b^2 +c^2=48$. Prove that:

$$a^2\sqrt{2b^3+16}+ b^2 \sqrt{2c^3+16}+ c^2 \sqrt{2a^3+16}\leq 24^2$$

Solution (G. Bas)



According to the Cauchy - Schwartz inequality we have that:

$$\begin{aligned}\left[\sum a^2\sqrt{2b^3+16}\right]^2&\leq\sum a^2(2b+4)\cdot\sum\frac{a^2(2b^3+16)}{2b+4}\\&=\left(2\sum a^2b+4\sum a^2\right)\cdot\sum\frac{a^2(2b+4)(b^2-2b+4)}{2b+4}\\&=\left(2\sum a^2b+192\right)\cdot\sum [a^2(b^2-2b+4)]\\&=\left(2\sum a^2b+192\right)\left(\sum a^2b^2-2\sum a^2b+192\right)\end{aligned}$$

Using the inequality $xy\leq\frac{1}{4}(x+y)^2$ as well as the (basic) inequality $ab+bc+ca\leq \frac{1}{3}(a+b+c)^2$ we have that:

$$\begin{aligned}\left(2\sum a^2b+192\right)\left(\sum a^2b^2-2\sum a^2b+192\right)&\leq\frac{1}{4}\left(\sum a^2b^2+2\cdot 192\right)^2\\&\leq\frac{1}{4}\cdot\left[\frac{1}{3}\left(\sum a^2\right)^2+2\cdot 192\right]^2\\&=\frac{1}{4}\cdot\left[\frac{48^2}{3}+2\cdot 192\right]^2\\&=(192+8\cdot 48)^2\\&=24^4\end{aligned}$$

ending the exercise.

The exercise can also be found in mathematica.gr