Proving an inequality involving fractions and square roots holds

Question

I have tried to prove the following inequality holds using a few approaches but none have worked. I am not really sure if I'm missing something.

Here's the question:

For every $x, y > 0$ prove that $\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}\ge\sqrt{x}+\sqrt{y}$

I have tried squaring both sides and re-arranging the inequality which led me to $(\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y})(\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}+\sqrt{x}+\sqrt{y}) \ge 0$, which doesn't help much.

I also tried finding a common denominator which led me to $$\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{xy}} \ge \sqrt{x}+\sqrt{y}$$ $$x\sqrt{x}+y\sqrt{y}\ge x\sqrt{y}+y\sqrt{x}$$This also wasn't a successful attempt.

I likewise thought about splitting the proof into two cases: $x\ge y$ and $y\ge x$ but I'm not sure how to proceed if I will try to prove it this way.

Could someone lend me a hand? Thanks a lot in advance!

Answer 1

We need to prove that $$\sqrt{x^3}+\sqrt{y^3}\geq(\sqrt{x}+\sqrt{y})\sqrt{xy}$$ or $$x-\sqrt{xy}+y\geq\sqrt{xy}$$ or $$(\sqrt{x}-\sqrt{y})^2\geq0.$$ Also, we can use Rearrangement:

$\left(\frac{1}{\sqrt{x}},\frac{1}{\sqrt{y}}\right)$ and $(x,y)$ have opposite ordering.

Thus, $$\frac{1}{\sqrt{x}}\cdot x+\frac{1}{\sqrt{y}}\cdot y\leq \frac{1}{\sqrt{y}}\cdot x+\frac{1}{\sqrt{x}}\cdot y,$$ which is our inequality.

Answer 2

Hint: Try to consider the 3 cases;

CASE 1: $x=y$ (equality holds)

CASE 2: $x<y$ (Hint:- $y/x >1$ & $\sqrt{\dfrac{y^2}{x}} =\sqrt{\dfrac{y.y}{x}} >\sqrt y $)

CASE 3: $x>y$ (This is trivial if you prove case 2)