Mathematical Reflections 2017, Issue 5 - Problem S421

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

Proposed by Constantinos Metaxas, Athens, Greece

Solution:
We prove the stronger inequality $$\dfrac{a^2}{\sqrt{1+a}}+\dfrac{b^2}{\sqrt{1+b}}+\dfrac{c^2}{\sqrt{1+c}} \geq \dfrac{3}{\sqrt{2}}.$$
Let $f(x)=\dfrac{x^2}{\sqrt{1+x}}$. Since $f''(x)=\dfrac{3x^2+8x+8}{4(x+1)^2\sqrt{x+1}}$, then $f''(x)>0$ for all $x>0$, so $f$ is convex on $(0,+\infty)$. By Jensen's Inequality, we have
$$f\left(\dfrac{a+b+c}{3}\right) \leq \dfrac{f(a)+f(b)+f(c)}{3},$$ i.e.
$$\dfrac{a^2}{\sqrt{1+a}}+\dfrac{b^2}{\sqrt{1+b}}+\dfrac{c^2}{\sqrt{1+c}} \geq 3\cdot \dfrac{\left(\frac{a+b+c}{3}\right)^2}{\sqrt{1+\frac{a+b+c}{3}}}=\dfrac{(a+b+c)^2}{\sqrt{9+3(a+b+c)}}.$$
By the AM-GM Inequality, we have $a+b+c \geq 3\sqrt[3]{abc}=3$. Set $x=a+b+c$. Observe that the function $g(x)=\dfrac{x^2}{\sqrt{9+3x}}$ is increasing on $[3,+\infty)$, so $g(x) \geq g(3)=\dfrac{3}{\sqrt{2}}$ and the conclusion follows.