# If $a>0, b>0$, prove that $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}} \geq \sqrt{a} + \sqrt{b}$

If $$a>0, b>0$$, prove that

$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}} \geq \sqrt{a} + \sqrt{b}.$$

I tried to multiply both sides by $$\sqrt{ab}$$ , but I only got this:

$$\sqrt{b}\geq\sqrt{a}.$$

I think it does not help at all. Could give me some hints, please.

## 7 Answers

Hint:   for $$\,a,b \gt 0\,$$, $$\,a \le b \iff \dfrac{1}{a} \ge \dfrac{1}{b} \iff \dfrac{1}{\sqrt{a}} \ge \dfrac{1}{\sqrt{b}}\,$$, so $$\,\left(a-b\right)\left(\dfrac{1}{\sqrt{a}}-\dfrac{1}{\sqrt{b}}\right) \le 0\,$$.

Or, use the rearrangement inequality directly.

• @ultrahamster Thank you for the suggested edit, but I prefer to leave it as a hint rather than a full answer. – dxiv Oct 2 '18 at 1:20

Hint: Try to find $$a$$ and $$b$$ sucht that

$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}<\sqrt{a}+\sqrt{b}.$$

OBS: This is not possible, then you prove your inequality.

Upon squaing both sides and taking common denominator, we get $$\frac{a}{\sqrt{b}}+\frac{b} {\sqrt{a}}\ge\sqrt{a}+\sqrt{b}$$

$$\iff a^3 +b^3 \ge ab ( a+b)$$ $$\iff a^2-ab + b^2 \ge ab$$ $$(a-b)^2\ge 0$$

The inequality is equivalent to $$\frac{(\sqrt{a} - \sqrt{b})^2 (\sqrt{a} + \sqrt{b})}{\sqrt{a} \sqrt{b}} \ge 0$$

You can simply take the difference between left and right sides of the inequality $$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}-(\sqrt{a}+\sqrt{b})= \frac{a-b}{\sqrt{b}}+\frac{b-a}{\sqrt{a}}= (a-b)\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{a}}\right)= \frac{(a-b)(\sqrt{a}-\sqrt{b})}{\sqrt{ab}}\geq 0$$

Let $$x=\sqrt{a}$$ and $$y=\sqrt{b}$$, then

\begin{align} \frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}} &= \frac{x^2}{y}+\frac{y^2}{x }\\ &= \frac{x^3+y^3}{xy} \\ &= \frac{(x+y)(x^2-xy+y^2)}{xy} \\ &= \frac{(x+y)[(x-y)^2+xy]}{xy} \\ & \ge \frac{(x+y)xy}{xy} \\ &= x+y \\ &= \sqrt{a}+\sqrt{b} \end{align}

You can multiply by $$\sqrt{ab}$$: $$\sqrt{ab}\cdot \left(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}\right) \geq \sqrt{ab}\cdot \left(\sqrt{a} + \sqrt{b}\right) \Rightarrow \\ a\sqrt{a}+b\sqrt{b}\ge a\sqrt{b}+b\sqrt{a} \Rightarrow \\ \sqrt{b}(b-a)\ge \sqrt{a}(b-a).$$ However, you can not divide by $$b-a$$, because you do not know if it is positive or negative (it effects the inequality sign). Instead, you can move the RHS to LHS: $$\sqrt{b}(b-a)- \sqrt{a}(b-a)\ge 0 \Rightarrow \\ (b-a)(\sqrt{b}-\sqrt{a})\ge 0.$$ Now consider two cases: 1) $$b\ge a$$; 2) $$b. Can you finish?