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Assuming $x,y,z\in [-1,1]$, suppose that

$$1+2xyz\geqslant x^2 + y^2 + z^2$$

Can we infer from this that $1+2(xyz)^n\geqslant x^{2n} + y^{2n} + z^{2n}$ for any positive integer $n$?

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Seems true. But I can't prove it. – Patrick Li Jan 31 '13 at 16:24
if $x=y=z=-1$ then $1+2 \cdot (-1)(-1)(-1)=1-2=-1 \leq (-1)^2+(-1)^2+(-1)^2=3$ so it's not true. – Iuli Feb 4 '13 at 9:12

This is not true. Since the first inequality is not true.

Basically, take x=y=z=-0.9, then

1+2xyz = - 0.458 < 0.729 = x^2 + y^2 + z^2.

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I don't think that's what the OP is asking. Rather: Given the fact that the first inequality holds, does the second one hold too? – mrf Feb 4 '13 at 9:06

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