Solving Inequality $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$ If someone could help with solving the inequality above, that would be awsome!
Here is my thinking of using AGM

*

*$\sqrt{\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)}\le \frac{\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}}{2}$

*$\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)\le \frac{\left(x\sqrt{x}+y\sqrt{y}\right)^2}{4xy}$           (Square both sides)

*$\sqrt{x}\sqrt{y}\le \:\frac{\left(x\sqrt{x}+y\sqrt{y}\right)^2}{4xy}$ (Simplify Left Side)

*$4xy\sqrt{x}\sqrt{y}\le x^3+2xy\sqrt{x}\sqrt{y}+y^3$ (Move 4xy to the other side and Square remaining)

*But now I see that this will not end up in $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$
Please help????

Let $u = \sqrt{x/y}$. The problem is equivalent to showing
$$
u^3 + 1 \ge u^2 + u,
$$
which follows immediately from
$$
(u-1)^2(u+1) \ge 0
$$
for all $u \ge 0$.
