<s>[INST] 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?
 [/INST] There is a simpler way, by eliminating at once the square roots.
Setting $x=X^2$ and $y=Y^2$ with $X>0$ and $Y>0$, the issue is equivalent to:
$$\text{show that} \ \ \ \dfrac{X^2}{Y}+\dfrac{Y^2}{X} \geq X+Y$$
which amounts to prove that:
$$X^3+Y^3 \geq XY(X+Y)$$
knowing factorization $X^3+Y^3=(X+Y)(X^2-XY+Y^2)$ and simplifying by $X+Y$, one is brought back to show that $X^2-2XY+Y^2 \geq 0$ i.e. $(X-Y)^2\geq 0$ which is eviedently true.
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