algebra precalculus - Prove $sumlimits_{i=1}^{n}frac{a_{i}^{2}}{b_{i}} geq frac{(sumlimits_{i=1}^{n}a_i)^2}{sumlimits_{i=1}^{n}b_i}$

So I have the following problem, which I'm having trouble solving:

Let $a_1$ , $a_2$ , ... , $a_n$ be real numbers. Let $b_1$ , $b_2$ , ... , $b_n$ be positive real numbers. Prove

$$ \frac{a_{1}^{2}}{b_{1}} + \frac{a_{2}^{2}}{b_{2}} + \cdot \cdot \cdot +\frac{a_{n}^{2}}{b_{n}} \geq \frac{(a_{1}+a_{2}+\cdot \cdot \cdot+a_{n})^2}{b_{1}+b_{2}+\cdot \cdot \cdot+b_{n}} $$

I was thinking that I somehow could use the Cauchy–Schwarz inequality, but with no success.

Any help would be very appreciated

Answer

You can simply use the cauchy-scwartz on the sets $$\left\{\frac{a_1}{\sqrt {b_1}},\frac{a_2}{\sqrt {b_2}},\dots,\frac{a_n}{\sqrt {b_n}}\right\}\text{ and }\left\{\sqrt {b_1},\sqrt {b_2},\dots,\sqrt {b_n}\right\}.$$

What you will get, is $$\left(\frac{a_{1}^{2}}{b_{1}} + \frac{a_{2}^{2}}{b_{2}} + \cdot \cdot \cdot +\frac{a_{n}^{2}}{b_{n}}\right)(b_{1}+b_{2}+\dots+b_{n}) \geq (a_{1}+a_{2}+\dots+a_{n})^2.$$