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?

Your proof is valid, but I suggest an alternative.
Let $a=\sqrt{x}>0$ and $b=\sqrt{y}>0$. You want to show
$$
\frac{a^2}{b}+\frac{b^2}{a}\ge a+b
$$
which becomes
$$
\frac{a^3+b^3}{ab}\ge a+b
$$
or
$$
\frac{(a+b)(a^2-ab+b^2)}{ab}\ge a+b
$$
Since $a+b>0$ and $ab>0$, we can factor out $a+b$ and remove the denominator, so the inequality becomes
$$
a^2-ab+b^2\ge ab
$$
Can you finish?

Your strategy works as well: the inequality is equivalent to
$$
\frac{a^2}{b}+\frac{b^2}{a}-a-b\ge0
$$
that becomes
$$
\frac{a^3+b^3-ab(a+b)}{ab}\ge0
$$
The numerator can be rewritten as
$$
a^3+b^3-ab(a+b)=(a+b)(a^2-ab+b^2)-ab(a+b)=
(a+b)(a-b)^2
$$
So the inequality is
$$
\frac{(a+b)(a-b)^2}{ab}\ge0
$$
which is true.
