$a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ $a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ (use Cauchy–Schwarz inequality)
I have trouble finding the two vectors. Is it $(x,y,z)$ and $(a,b,c)$?
I need a hint

By C-S $$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=\frac{\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)(a+b+c)}{a+b+c}\geq$$
$$\geq\frac{\left(\frac{x}{\sqrt{a}}\cdot\sqrt{a}+\frac{y}{\sqrt{b}}\cdot\sqrt{b}+\frac{z}{\sqrt{c}}\cdot\sqrt{c}\right)^2}{a+b+c}=\frac{(x+y+z)^2}{a+b+c}.$$
The equality occurs for $$\left(\frac{x}{\sqrt{a}},\frac{y}{\sqrt{b}},\frac{z}{\sqrt{c}}\right)||(\sqrt{a},\sqrt{b},\sqrt{c})$$ or
$$(x,y,z)||(a,b,c).$$
Actually, the last writing is very useful. 
