<s>[INST] 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. 
 [/INST] 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
$$\text{LHS} = ax+by+cz + dt \le \sqrt{(a^2+b^2+c^2+d^2)(x^2+y^2+z^2+t^2)} = a+b+c = \text{RHS}$$
</s>