$x^2+y^2+\frac8{xy}\ge2xy+\frac8{xy}\ge8$ by AM-GM. Equality when $x=y=\sqrt2$.

By AM-GM, $$\frac{x^2+\frac{4}{xy}+\frac{4}{xy}+y^2}{4} \geq \sqrt[4]{x^2 \cdot \frac{4}{xy} \cdot \frac{4}{xy} \cdot y^2}$$ $$x^2+\frac{8}{xy}+y^2 \geq 8$$

Prove that $$x^2 +\frac{k}{xy}+ y^2 \ge 2\sqrt{2k}$$for all positive real numbers $x,y$ and $k.$

Prove that $$x +\frac{8}{xy}+ y \ge 6$$$$x +\frac{8}{xy}+ y^2 \ge 5\sqrt[5]{4}$$for all positive real numbers $x$ and $y$.

$x^2+y^2+\frac k{xy}\ge2xy+\frac k{xy}\ge2\sqrt{2k}$ $x+\frac8{xy}+y\ge6$ directly by AM-GM. $\frac12x+\frac12x+\frac4{xy}+\frac4{xy}+y^2\ge5\sqrt[4]5$

Let $x,y\geq \frac{1}{2}.$Prove that $$x +y+\frac{4}{xy}-\frac{1}{y} \ge 4$$$$x +y+\frac{2}{xy}-\frac{1}{y} \ge \frac{5}{2} $$$$x +y+\frac{3}{xy}-\frac{1}{y} \ge 2\sqrt 6-\frac{3}{2} $$$$x +y+\frac{1}{xy}-\frac{1}{x}-\frac{1}{y} \ge \frac{1}{2}$$$$x +y+\frac{1}{2xy}-\frac{1}{x}-\frac{1}{y} \ge-1$$$$x +y+\frac{3}{2xy}-\frac{1}{x}-\frac{1}{y} \ge 2\sqrt 2-\frac{3}{2}$$$$x +y+\frac{2}{xy}-\frac{1}{x}-\frac{1}{y} \ge 2\sqrt 3-\frac{3}{2}$$

sqing wrote: Prove that $$x +y+\frac{4}{xy}-\frac{1}{y} \ge 4$$$$x +y+\frac{2}{xy}-\frac{1}{x}-\frac{1}{y} \ge 2$$for all positive real numbers $x$ and $y$. $(x,y)=(5,0.1)$ is a counterexample to both.