$x+y+z+4xyz=2$ prove that$ xy+yz+xz\le 1$

Question

Let $x$,$y$,$z$ be non-negative real numbers, for which: $x + y + z + 4xyz=2$ Prove that: $xy + yz + zx \le1$. I am sure this can be done with Cauchy-Schwarz or AM-GM, but I have long forgotten how to use those...Any help is appreciated !

Answer 1

I don't know about using the inequalities, but this can be solved without any tricks!

\begin{eqnarray} (1-2x)(1-2y)(1-2z) & \leq & 1 \\ 1 - 2(x+y+z) + 4(xy+xz+yz)-8xyz & \leq & 1 \\ 1 - 2(x+y+z+4xyz) + 4(xy+xz+yz) & \leq & 1 \\ 1 - 4 + 4(xy+xz+yz) & \leq & 1 \\ xy+xz+yz & \leq & 1 \end{eqnarray}

The first inequality uses the fact that $x,y,z\geq 0$, and the fourth line substitutes the $x+y+z+4xyz$ from the third line for $2$.

Using $(1+2x)(1+2y)(1+2z)\geq 1$, one can similarly show that $xy+yz+zx\geq-1$

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Great point from the comments. The first inequality is not justified, and, in fact, there is an easy counter example to this. For example $x,y=1$ and $z=0$ satisfies the equation $x+y+z+4xyz=2$, yet $xy+xz+yz = 2\not\leq 1$. The second inequality that I gave above holds, and the first one would hold if, for example, $0\leq x,y,z\leq 0.5$ (lesser restrictions are also possible).