National Olympiad of Mathematics National Stage, Deva, April 23, 2019 , CLASS Xth Problem No. 1. Let a, b, c real numbers strictly positive. Show it as $$\frac{1}{abc}+1\ge3\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{a+b+c}\right) .$$Solution. Using inequalities $a^2 + b^2 + c^2 \geq 3 \sqrt[3]{a^2b^2c^2}$ and $a + b + c \geq 3 \sqrt[3]{abc} $..............................................2p $\ $ get $3\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{a+b+c}\right) \leq \frac{1}{\sqrt[3]{a^2b^2c^2}}+\frac{1}{\sqrt[3]{abc}}$...................................................................................... 2p $ \ $ $\ $ Noting $\frac{1}{\sqrt[3]{abc}}= t$, it's enough to look $t^3 + 1 \geq \ t^2 + t$ $ \ $ or $(t - 1)^2 (t + 1) \geq 0$, what is obvious. $\ $ ........................................................................................................................................................................................ 3p