Mathematical Reflections 2017, Issue 3 - Problem S412

Problem:
Let $a,b,c$ be positive real numbers such that
$$\dfrac{1}{\sqrt{1+a^3}}+\dfrac{1}{\sqrt{1+b^3}}+\dfrac{1}{\sqrt{1+c^3}} \leq 1.$$ Prove that $a^2+b^2+c^2 \geq 12$.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Observe that for any positive real number $x$ we have
$$\dfrac{1}{1+\frac{1}{2}x^2} \leq \dfrac{1}{\sqrt{1+x^3}} \iff 1+x^3 \leq \left(1+\dfrac{1}{2}x^2\right)^2 \iff x^2(x-2)^2 \geq 0.$$
It follows that
$$\sum_{cyc} \dfrac{1}{1+\frac{1}{2}a^2} \leq  \sum_{cyc} \dfrac{1}{\sqrt{1+a^3}} \leq 1.$$
By Cauchy-Schwarz Inequality, we get
$$\begin{array}{lll} \displaystyle 1\cdot\left(3+\dfrac{1}{2}(a^2+b^2+c^2)\right) & \geq & \displaystyle \sum_{cyc} \dfrac{1}{1+\frac{1}{2}a^2}\sum_{cyc} \left(1+\dfrac{1}{2}a^2\right) \\ & \geq & 9, \end{array}$$
so $a^2+b^2+c^2 \geq 12$.