Proof of the Bergström inequality using Cauchy

Question

$\sum_{i=1}^n \frac{a_i^2}{b_i} \geq \frac{(a_1+...+a_n)^2}{b_1+...+b_n}$

for real $a_i, b_i > 0$.

I'm told it follows from Cauchy-Schwarz, but I just don't see it.

Answer 1

Write the inequality as

$$\left(\sum_{i=1}^n a_i\right)^2 \leqslant \left(\sum_{i=1}^n b_i\right)\left(\sum_{i=1}^n \frac{a_i^2}{b_i}\right).$$

In that form, it is easier to see what to take for $x_i$ and $y_i$ in

$$\left(\sum_{i=1}^n x_iy_i\right)^2 \leqslant \left(\sum_{i=1}^n x_i^2\right)\left(\sum_{i=1}^n y_i^2\right).$$