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. 
 A: Let $d = \sqrt{2(ab+bc+ca)}$ and $t = \sqrt{2(xy+yz+xz)}$. It is easy to check
$$
\begin{cases}
a^2 + b^2 + c^2 + d^2 &= a^2+b^2+c^2 + 2(ab+bc+ca) = (a+b+c)^2\\
x^2 + y^2 + z^2 + t^2 &= x^2+y^2+z^2 + 2(xy+yz+xz) = (x+y+z)^2 = 1
\end{cases}
$$
Apply Cauchy Schwarz to the two 4-vectors $(a,b,c,d)$ and $(x,y,z,t)$, 
we immediately get
$$\text{LHS} = ax+by+cz + dt \le \sqrt{(a^2+b^2+c^2+d^2)(x^2+y^2+z^2+t^2)} = a+b+c = \text{RHS}$$
A: Let $a+b+c=k$.
Hence, $(a-kx)^2+(b-ky)^2+(c-kz)^2\geq0$ gives
$(a-kx)^2+(b-ky)^2+(c-kz)^2\geq(a-kx+b-ky+c-kz)^2$ or
$(xy+xz+yz)k^2-(ay+bx+az+cx+bz+cy)k+ab+ac+bc\geq0$, which gives
$4(ab+ac+bc)(xy+xz+yz)\leq(ay+bx+az+cx+bz+cy)^2$ or
$ax+by+cz+2\sqrt{(ab+ac+bc)(xy+xz+yz)}\leq a+b+c$. Done!
A: 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. 
A: WLOG: $$a+ b+ c= 1$$
AM-GM: 
$$ax+ by+ cz+ 2\sqrt{\left ( xy+ yz+ zx \right )\left ( ab+ bc+ ca \right )}$$
$$\leq  ax+ by+ cz+ xy+ yz+ zx+ ab+ bc+ ca$$
$$\Leftarrow  xy+ yz+ zx+ ab+ bc+ ca= \frac{1- x^{2}- y^{2}- z^{2}}{2}+ \frac{1- a^{2}- b^{2}- c^{2}}{2}$$
$$\leq  1- ax- by- cz$$
$$\Leftrightarrow  \left ( x- a \right )^{2}+ \left ( y- b \right )^{2}+ \left ( z- c \right )^{2}\geq  0$$
