Find maximum of $C=2(x+y+z)-xy-yz-xz$

Question

Let $x,y,z\ge 0$ such that $x^2+y^2+z^2=3$. Find the maximum of $$C=2(x+y+z)-xy-yz-xz.$$

I tried Schur and AM-GM inequality but I really have no idea about this problem. It is not homogeneous so it's hard for me.

Answer 1

Let $(x+y+z) = a$ . Then $$(x+y+z)^2 = a^2 \implies (xy+yz+zx)=\dfrac{a^2-3}{2}$$

So $$2(x+y+z)-xy-yz-xz = 2a -\dfrac{a^2-3}2$$

$$ C = \dfrac{-a^2+4a+3}{2}$$

The maximum of this quadratic is at $a = 2$ , for which the max becomes :

$$2(x+y+z)-xy-yz-xz = \dfrac 72$$