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 !
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See my updated answer. It turns out that the result is not true. – Dr Xorile Nov 16 '16 at 17:14
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There is a mistake in your updated answer: $x,y=1,z=0$ $\implies$ $xy+yz+zx=1$. – George Law Nov 17 '16 at 16:09
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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 $xy+yz+zx\geq-1$
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).
Dr Xorile
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I think $(1-2x)(1-2y)(1-2z)\leq 1$ needs better justification. $x,y,z\geq 0$ alone doesn't seem enough. If nonnegativity alone were enough, then consider $x=2$, $y=2$ and $z=0$. – yurnero Nov 16 '16 at 02:09
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Assuming WLOG that $x\ge y\ge z\ge0$ the inequality $(1-2x)(1-2y)(1-2z)\ge1$ would hold unless $y\ge\frac12>z$, where it is non-negative but not guaranteed to be less than or equal to 1. – George Law Nov 17 '16 at 17:57