Prove that $ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$ Let $a,b,c,x,y,z$ be positive real numbers such that $x+y+z=1$. Prove that 
$$ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$$.
my try: 
$2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{\frac{2(a+b+c)}{3}}$
But this is not the right choice because 
$ax+by+cz\le{\frac{a+b+c}{3}}$ is not always true. 

WLOG: $$a+ b+ c= 1$$
AM-GM: 
$$ax+ by+ cz+ 2\sqrt{\left ( xy+ yz+ zx \right )\left ( ab+ bc+ ca \right )}$$
$$\leq  ax+ by+ cz+ xy+ yz+ zx+ ab+ bc+ ca$$
$$\Leftarrow  xy+ yz+ zx+ ab+ bc+ ca= \frac{1- x^{2}- y^{2}- z^{2}}{2}+ \frac{1- a^{2}- b^{2}- c^{2}}{2}$$
$$\leq  1- ax- by- cz$$
$$\Leftrightarrow  \left ( x- a \right )^{2}+ \left ( y- b \right )^{2}+ \left ( z- c \right )^{2}\geq  0$$
