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?
 A: If $x$ and $y$ are assumed to be positive, then I do not see anything wrong, but I think it would be much better if you rewrote it from the bottom up instead of as you have it now (working from what you are trying to prove as opposed to to it, if that makes sense). 
A: 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 $x, y, z > 0 \implies \dfrac{x^2}{y} + \dfrac{y^2}{z} + \dfrac{z^2}{x} \geq x+y+z$.
A: Use Titus lemma:
$$\frac{x^2}{a}+\frac{y^2}b\ge \frac{(x+y)^2}{a+b}$$
Then $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}=\frac{\sqrt{x}^2}{\sqrt{y}}+\frac{\sqrt{y}^2}{\sqrt{x}}\ge \frac{(\sqrt{x}+\sqrt{y})^2}{(\sqrt{x}+\sqrt{y})}=\sqrt{x}+\sqrt{y}$$
A: First you neeed to qualify your starting point by saying that $x,y > 0$ and that by $\sqrt{x}$ or $\sqrt{y}$ the proposition means the positive value of that square root.  But these are tirival details.
The second flaw is that what you presented is not a proof; the very first line of the "proof" asserts the proposition that was to have been proven.  The steps you present are a very good way to discover the steps of the proof, but the proof itself would have to start with your last step and work its way upwards.  Thus:
$$(\sqrt{x}-\sqrt{y})^2 \geq 0 \wedge (\sqrt{x}+\sqrt{y}) \geq 0 \wedge \sqrt{y}\sqrt{x} > 0 \implies 
\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$
should have been the first step, not the last.  From there you go to your next-to-last step, and onward till you have reached the proposition to be proven.
A: 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.
A: 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.
