zschess wrote: Let $ a, b, c $ be positive real numbers less than or equal to $ \sqrt{2} $ such that $ abc = 2 $, prove that $$ \sqrt{2}\displaystyle\sum_{cyc}\frac{ab + 3c}{3ab + c} \ge a + b + c $$ $\frac{2 + 3c^2}{6 + c^2} \geq \frac{\sqrt{2}c}{2}\iff (\sqrt{2}-c)^3\geq0,$ $\displaystyle\sum_{cyc}\frac{ab + 3c}{3ab + c} =\displaystyle\sum_{cyc}\frac{2 + 3c^2}{6 + c^2} \geq \displaystyle\sum_{cyc}\frac{\sqrt{2}c}{2}=\frac{\sqrt{2}}{2}(a + b + c ).$

$\color{red}{\textit{\textbf{Proof:}}}$ Since \[\sqrt 2\sum_{cyc} \frac{ab+3c}{3ab+c} =\sqrt 2\sum_{cyc} \frac{2+3a^2}{6+a^2}\]and the function $f(x)=\frac{2+3x^2}{6+x^2}$ has its second derivative as $f''(x)=\frac{96(2-x^2)}{(x^2+6)^3} \ge 0$ for all positive real $x\le \sqrt 2$, $f(x)$ is convex which by Jensen's Inequality, we get \[\sqrt 2(f(a)+f(b)+f(c))\ge 3\sqrt 2f\left(\frac{a+b+c}{3}\right) \ge a+b+c\]if and only if \[\frac{3\sqrt 2 (2+\frac{s^2}{3})}{6+\frac{s^2}{9}}\ge s\]where $s=a+b+c$ which is true when $s\le 3\sqrt 2$. $\quad \blacksquare$