prove that this equality is always right for each positive x and y. prove that this inequality is hold for each positive x,y. 
$x\over\sqrt{y}$ + $y\over\sqrt{x}$  $\ge$ $\sqrt{x}$ + $\sqrt{y}$  
I want a detailed way of solving the question. 
 A: This is purely algebraic... You showed that the inequality to prove is equivalent to $$x\sqrt{x}+y\sqrt{y}\geqslant x\sqrt{y}+y\sqrt{x},$$ or, equivalently, $$(\sqrt{x}+\sqrt{y})(x-\sqrt{xy}+y)\geqslant (\sqrt{x}+\sqrt{y})\sqrt{xy},$$ hence it suffices to show that $$x-\sqrt{xy}+y\geqslant\sqrt{xy}.$$ Surely you can show that  $$x-2\sqrt{xy}+y\geqslant0...$$ Shortcut: $$x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{y}+y\sqrt{x}\right)=(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})^2\geqslant0.$$ Or again, starting from the formulation you were given: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}=\sqrt{x}+\sqrt{y}+\left(\frac1{\sqrt{x}}+\frac1{\sqrt{y}}\right)(\sqrt{x}-\sqrt{y})^2\geqslant\sqrt{x}+\sqrt{y}.$$
A: Let $u=\sqrt{x},\ v=\sqrt{y}$ so it becomes $u^2/v+v^2/u \ge u+v.$ Now move the right side to the left and factor. The result is
$$\frac{(u+v)(u-v)^2}{uv}.$$
Since this is $\ge 0$ the inequality holds.
A: Hint one way is to use Cauchy Schwarz inequality, it is quite direct. Try it, and in case you have trouble will expand. 
A: Without loss of generality you can assume $0<x<y$. Then use the rearrangement inequality:
$$
\sqrt x + \sqrt y = \frac{x}{\sqrt x} + \frac{y}{\sqrt y} \le 
\frac{x}{\sqrt y} + \frac{y}{\sqrt x}
$$
