Selection Examination

(α) We have $$\begin{array}{rl}{x}^3-3x\ge -2& \iff x^3-3x+2\ge 0\\ & \iff x^3-x-2x+2\ge 0\\ & \iff x(x-1)(x+1)-2(x-1)\ge 0\\ & \iff (x-1)(x^2+x-2)\ge 0\\ & \iff (x+2)(x-1{)}^2\ge 0\end{array}$$ which is valid, because $x>0$.

(β) The inequality is equivalent to $$\frac{y}{z}\left(x^2+\frac{2}{x}\right)+\frac{z}{x}\left(y^2+\frac{2}{y}\right)+\frac{x}{y}\left(z^2+\frac{2}{z}\right)\ge 9.(2)$$ From question (α), by dividing both parts by $x$ we obtain $x^2+\frac{2}{x}\ge 3$. Therefore it is enough to prove: $$3\left(\frac{y}{z}+\frac{z}{x}+\frac{x}{y}\right)\ge 9\iff \frac{y}{z}+\frac{z}{x}+\frac{x}{y}\ge 3.(3)$$ The latter is valid, as we can easily see, by applying the inequality of arithmetic and geometric mean as follows: $$\frac{y}{z}+\frac{z}{x}+\frac{x}{y}\ge 3\sqrt{\frac{yzx}{zxy}}=3(4)$$ The equality holds if and only if $x=y=z=1$.