This problem was in IMC 2010.

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 $n$?

  • 1
    $\begingroup$ Seems true. But I can't prove it. $\endgroup$
    – JACKY88
    Jan 31, 2013 at 16:24
  • $\begingroup$ 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. $\endgroup$
    – Iuli
    Feb 4, 2013 at 9:12

1 Answer 1


We can assume that $ab\geq0$ because

if $ab<0$, $ac<0$ and $bc<0$ we get $a^2b^2c^2<0$, which is contradiction.

The condition gives $(1-a^2)(1-b^2)\geq(c-ab)^2$ and we need to prove that $$\left(1-a^{2n}\right)\left(1-b^{2n}\right)\geq\left(c^n-a^nb^n\right)^2,$$ which is C-S: $$\left(c^n-a^nb^n\right)^2=(c-ab)^2\left(\sum\limits_{i=0}^{n-1}c^i(ab)^{n-1-i}\right)^2\leq(c-ab)^2\left(\sum\limits_{i=0}^{n-1}(ab)^{n-1-i}\right)^2\leq$$ $$\leq(1-a^2)(1-b^2)\sum_{i=0}^{n-1}a^{2i}\sum_{i=0}^{n-1}b^{2i}=\left(1-a^{2n}\right)\left(1-b^{2n}\right)$$ and we are done!


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