How to prove that $\frac x{\sqrt y}+\frac y{\sqrt x}\ge\sqrt x+\sqrt y$

Question

I am trying to prove that $$\frac x{\sqrt y}+\frac y{\sqrt x}\ge\sqrt x+\sqrt y$$ I tried some manipulations, like multiplications in $\sqrt x$ or $\sqrt y$ or using $x=\sqrt x\sqrt x$, but I'm still stuck with that.

What am I missing?

Answer 1

Diviging by $\sqrt y$ we get,

$$\frac x{y}+\frac{\sqrt y}{\sqrt x}\ge\sqrt{\frac xy}+1$$ which can be written $$t^2+\frac1t\ge t+1$$ or $$t^3-t^2-t+1=(t-1)^2(t+1)\ge0$$ for $t>0$. Equality occurs with $t=1$, i.e. $x=y$.

Answer 2

Let $a>0$, $b>0$, we have that \begin{align} \frac{a^2} b+\frac{b^2} a\geq a+b&\iff \frac{a^3+b^3} {ab}\geq a+b\\ &\iff a^2-ab+b^2\geq ab\\ &\iff (a-b)^2\geq0 \end{align}

Answer 3

Let $x\geq y$.

Hence, $(x,y)$ and $\left(\frac{1}{\sqrt{x}},\frac{1}{\sqrt{y}}\right)$ are opposite ordered.

Thus, by Rearrangement $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq\frac{x}{\sqrt{x}}+\frac{y}{\sqrt{y}}=\sqrt{x}+\sqrt{y}$ and we are done!