$x^2+xy+xz+yz=6+2\sqrt{5}$. Find the minimum of $3x+y+2z$ Let $x,~y,~z>0$ satisfy $x^2+xy+xz+yz=6+2\sqrt{5}$. How to find the minimum of $3x+y+2z$? I can find it by Larange multiplier in calculus, but I wonder if there is a easy way using, say Cauchy-Schwartz inequality?

Apply the AM-GM inequality
$$3x+y+2z = (x+y)+2(x+z) \ge 2\sqrt{2(x+y)(x+z)} = 2\sqrt{2(x^2+xy+xz+yz)}=2\sqrt{2}(\sqrt{5}+1)$$
The equality occurs iff $x+y = 2(x+z)= \sqrt{6+2\sqrt{5}}$ or
$$x+y = 2(x+z) = \sqrt{5}+1$$
