Prove that that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \geq \frac{(x+y+z)^2}{a+b+c}.$ 
Prove that that $\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c} \geq \dfrac{(x+y+z)^2}{a+b+c}.$ with $a,b,c$ positive real numbers.

Attempt
I tried using Cauchy-Schwarz, but I can't find the correct $a_i$ and $b_i$. How would you solve this using Cauchy-Schwarz? 
 A: More generally, the following is called Titu's Lemma or Engel's form of Cauchy-Schwarz inequality:
For all $a_i\in\mathbb R$, $b_i\in\mathbb R^+$:
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\cdots+a_n)^2}{b_1+b_2+\cdots+b_n}$$
Proof: by Cauchy-Schwarz: $$(b_1+b_2+\cdots+b_n)\left(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\right)$$
$$\ge \left(\sqrt{b_1}\sqrt{\frac{a_1^2}{b_1}}+\sqrt{b_2}\sqrt{\frac{a_2^2}{b_2}}+\cdots+\sqrt{b_1}\sqrt{\frac{a_n^2}{b_n}}\right)^2$$
$$=(|a_1|+|a_2|+\cdots+|a_n|)^2\ge (a_1+a_2+\cdots+a_n)^2$$
with equality if and only if $\frac{a_1^2}{b_1^2}=\frac{a_2^2}{b_2^2}=\cdots=\frac{a_n^2}{b_n^2}$ and $(|a_1|+|a_2|+\cdots+|a_n|)^2= (a_1+a_2+\cdots+a_n)^2$, i.e. if and only if $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots=\frac{a_n}{b_n}$.
A: HINT: the left-hand side minus the right-hand side is equal to
$${z}^{2}{a}^{2}b+{y}^{2}{a}^{2}c+{z}^{2}a{b}^{2}-2\,abcxy-2\,abcxz-2\,a
bcyz+{y}^{2}a{c}^{2}+{x}^{2}{b}^{2}c+{x}^{2}b{c}^{2}
\geq 0$$
can you proceed? (sum of squares!)
