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. 

By Cauchy-Schwarz,
$(LHS)\le \sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}+\sqrt{2(ab+bc+ca)2(xy+yz+zx)}=f(a,b,c,x,y,z)$
However, $$f(a,b,c,x,y,z)^2\le (a^2+b^2+c^2+2ab+2bc+2ca)(x^2+y^2+z^2+2xy+2yz+2zx)=(RHS)^2(\because Cauchy)$$  
Therefore, $(LHS)^2\le (RHS)^2$. Our proof is done. 
