Prove: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$ 
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive

$$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$
$$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$
Is the proof valid?

Alternatively, apply AM-GM inequality with $a = \sqrt{x}, b = \sqrt{y}$:
$\dfrac{a^2}{b} + b \geq 2\sqrt{\dfrac{a^2}{b}\cdot b} = 2a, \dfrac{b^2}{a} + a \geq 2\sqrt{\dfrac{b^2}{a}\cdot a} = 2b$. Adding these $2$ inequalities to get the answer.
Generalization: The same technique could be used to prove: If $x, y, z > 0 \implies \dfrac{x^2}{y} + \dfrac{y^2}{z} + \dfrac{z^2}{x} \geq x+y+z$.
