parmenides51 wrote: a) Prove that for $x,y \ge 1$, holds $$x+y - \frac{1}{x}- \frac{1}{y} \ge 2\sqrt{xy} -\frac{2}{2\sqrt{xy}}$$ b) Prove that for $a,b,c,d \ge 1$ with $abcd=16$ , holds $$a+b+c+d-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}-\frac{1}{d}\ge 6$$ https://artofproblemsolving.com/community/c6h1828055p12232572

The inequality (a) perhaps has a typo. Should be $x+y-\frac{1}{x}-\frac{1}{y} \ge 2 \sqrt{xy}-\frac{2}{\sqrt{xy}}$, which holds because by re-arranging terms: \begin{align*} & \implies x+y- 2 \sqrt{xy} \ge \frac{1}{x}+\frac{1}{y}-\frac{2}{\sqrt{xy}} \\ & \implies (\sqrt{x}-\sqrt{y})^2 \ge \left( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{y}} \right)^2 \\ & \implies (\sqrt{x}-\sqrt{y})^2 \ge \frac{(\sqrt{y}-\sqrt{x})^2}{x \cdot y} \\ & \implies 1 \ge \frac{1}{x \cdot y} \\ & \implies x \cdot y \ge 1 \end{align*} which holds

Let $ x,y \ge 1 .$ Prove that $$x+y+2xy - \frac{1}{x}- \frac{1}{y} \ge \left(2+ 3\sqrt{3}\right)\left(\sqrt{xy} -\frac{1}{\sqrt{xy}}\right)$$

indeed (a) had a typo, the correct is $\frac{2}{\sqrt{xy}}$ instead of $\frac{2}{2\sqrt{xy}}$