Titu's Lemma
Titu's lemma states that for positive real numbers $a_1,\ a_2,\ ...,\ a_n$ and $b_1,\ b_2,\ ...,\ b_n$:
$\dfrac{a_1^2}{b_1} + \dfrac{a_2^2}{b_2} +... + \dfrac{a_n^2}{b_n} \ge \dfrac{(a_1 + a_2 + ... + a_n)^2}{b_1 + b_2 + ... + b_n}$
From Cauchy Schwarz inequality explained , we know that:
$(x_1^2 + x_2^2 + ... + x_n^2) (y_1^2 + y_2^2 + ... + y_n^2) \ge (x_1y_1 + x_2y_2 + ... + x_ny_n)^2$
We substitute each $x_i = \dfrac{a_i}{\sqrt{b_i}}$ and $y_i = {\sqrt{b_i}}$, in the Cauchy Schwarz inequality and get:
$\left\{\left(\dfrac{a_1}{\sqrt{b_1}}\right)^2 + \left(\dfrac{a_2}{\sqrt{b_2}}\right)^2 + ... + \left(\dfrac{a_n}{\sqrt{b_n}}\right)^2\right\}\times$$\ \ \left\{(\sqrt{b_1})^2 + (\sqrt{b_2})^2 + ... + (\sqrt{b_n})^2\right\} \ge $$\ \ \left\{\left(\dfrac{a_1}{\sqrt{b_1}}\right)\sqrt{b_1} + \left(\dfrac{a_1}{\sqrt{b_1}}\right)\sqrt{b_1} + ... + \left(\dfrac{a_1}{\sqrt{b_1}}\right)\sqrt{b_1}\right\}^2$
$\Rightarrow \left\{\dfrac{a_1^2}{b_1} + \dfrac{a_2^2}{b_2} + ... + \dfrac{a_n^2}{b_n}\right\}\left\{b_1 + b_2 + ... + b_n\right\} \ge \left\{a_1 + a_2 + ... + a_n\right\}^2$
$\Rightarrow \dfrac{a_1^2}{b_1} + \dfrac{a_2^2}{b_2} + ... + \dfrac{a_n^2}{b_n} \ge \dfrac{\left(a_1 + a_2 + ... + a_n\right)^2}{b_1 + b_2 + ... + b_n}$