Let $x,y,z\geq 0$. Then $2(x+y)(x+z)(y+z)+2xyz\le(xy+yz+xz)^2+(x+y+z)^2$.

Question

Suppose that $x,y,z\geq 0$ are three reals. I want to prove $$2(x+y)(x+z)(y+z)+2xyz\le(xy+yz+xz)^2+(x+y+z)^2.$$

My attempt: By AM-GM: $$LHS\le \frac2{27} (2x+2y+2z)^3+2xyz=\frac{16}{27}(x+y+z)^3+2xyz.$$

I am not sure what the right way to continue is.

Answer 1

Hint: It is $$(xy+yz+xz)^2-2(x+y)(x+z)(y+z)-2xyz+(x+y+z)^2=\left( xy+xz+yz-x-y-z \right) ^{2}$$

Answer 2

Hint: Since $a^2+b^2\geq2ab$ we have $$ (xy+yz+xz)^2+(x+y+z)^2\geq 2(xy+yz+xz)(x+y+z)$$

Now check if $$(x+y)(x+z)(y+z)+xyz \leq (xy+yz+xz)(x+y+z)$$

is true (multiply everything and cancel all you can).

Answer 3

Since $$(x+y)(x+z)(y+z)+xyz=(x+y+z)(xy+xz+yz),$$ we need to prove that $$2(x+y+z)(xy+xz+yz)\leq(x+y+z)^2+(xy+xz+yz)^2$$ or $$(x+y+z-xy-xz-yz)^2\geq0.$$