show that the maximum of $xy+ xz + yz$ is when $x = y = z$

Question

I have got to this while trying to prove something in combinatorics if someone has any solution for it, it will be very helpful

Answer 1

Hint

$$(a-b)^2=a^2+b^2-2ab \implies 2ab = a^2+b^2-(a-b)^2$$

so: $$2\left(xy+xz+yz\right) = 2x^2+2y^2+2z^2-\left(x-y\right)^2-\left(y-z\right)^2-\left(z-x\right)^2$$

Answer 2

Without any constraints on $x,y,z$, this is not really a sensible question since $xy + yz + zx$ can be arbitrarily large for stupid reasons. With the constraint $x,y,z\ge 0, x+y+z = C$ this follows from $$ xy + yz + zx = \frac{1}{3}(x+y+z)^2 - \frac{1}{6}\Big((x-y)^2 + (y-z)^2 + (z-x)^2\Big)\ . $$ On the other hand, if the constraint is, for example, $x,y,z\ge 0,xyz = C$, the arithmetic mean-geometric mean inequality shows that $xy + yz + zx$ is minimal for $x = y= z$.