Solving Inequality $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$ If someone could help with solving the inequality above, that would be awsome!
Here is my thinking of using AGM

*

*$\sqrt{\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)}\le \frac{\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}}{2}$

*$\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)\le \frac{\left(x\sqrt{x}+y\sqrt{y}\right)^2}{4xy}$           (Square both sides)

*$\sqrt{x}\sqrt{y}\le \:\frac{\left(x\sqrt{x}+y\sqrt{y}\right)^2}{4xy}$ (Simplify Left Side)

*$4xy\sqrt{x}\sqrt{y}\le x^3+2xy\sqrt{x}\sqrt{y}+y^3$ (Move 4xy to the other side and Square remaining)

*But now I see that this will not end up in $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$
Please help????
 A: Note that we must have $x > 0$ and $y > 0$ for the inequality to even be defined at all.
Start by squaring both sides to get the equivalent inequalty
$$\frac{x^2}{y} + \frac{y^2}{x} + 2 \sqrt{\frac{x^2}{y} \frac{y^2}{x}} = \frac{x^2}{y} + \frac{y^2}{x} + 2 \sqrt{xy} \geq x + y + 2 \sqrt{xy}$$
Equivalently,
$$\frac{x^2}{y} + \frac{y^2}{x} \geq x + y$$
Multiply both sides by $xy$ to get the equivalent inequality
$$x^3 + y^3 = (x + y)(x^2 - xy + y^2)\geq x^2 y + y^2 x = xy(x + y)$$
Divide both sides by $x + y$ to get the equivalent inequality $x^2 - xy + y^2 \geq xy$. Subtract $xy$ from both sides to get the equivalent inequality
$$(x - y)^2 = (x^2 - 2x + y^2) \geq 0$$
Which is clearly true.
A: Let $u = \sqrt{x/y}$. The problem is equivalent to showing
$$
u^3 + 1 \ge u^2 + u,
$$
which follows immediately from
$$
(u-1)^2(u+1) \ge 0
$$
for all $u \ge 0$.
A: Here's an elementary way to handle it. First put on the same denimator
$$\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}=\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{xy}}$$
then factor the numerator using the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$
$$\sqrt{x^3}+\sqrt{y^3}=(\sqrt{x}+\sqrt{y})(\sqrt{x^2}-\sqrt{xy}+\sqrt{y^2})$$
but
$$\frac{\sqrt{x^2}-\sqrt{xy}+\sqrt{y^2}}{\sqrt{xy}}=
\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}-1\geqslant 1$$
by the inequality $a+\frac{1}{a}\geqslant2$ for any $a>0$.
Alternative: use Cauchy-Schwarz inequality to prove that
$$\left(\frac{a^2}{b}+\frac{b^2}{a}\right)(b+a)\geqslant (a+b)^2$$
for all positive $a,b$, thus
$$\frac{a^2}{b}+\frac{b^2}{a}\geqslant a+b.$$
Now just take $x=a^2$ and $y=b^2$.
