<s>[INST] 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????
 [/INST] 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.
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