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Prove/disprove that $a^{2m} + b^{2m} + c^{2m} > 2^{1-m}$ subject to $a + b + c= 0$ and $a^2 + b^2 + c^2 = 1$

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Problem. Let $a, b, c$ be reals with $abc\ne 0$, $a + b + c = 0$, and $a^2 + b^2 + c^2 = 1$. Prove or disprove that$a^{2m} + b^{2m} + c^{2m} > 2^{1-m}, \forall m\in \mathbb{Z}_{>2}$.

Prior thoughts.

  • I saw a post here, where it was proved that if $a+b+c=0$ and $a^2+b^2+c^2=1$ then $a^4+b^4+c^4=1/2$. I was going to add a comment to that post but cannot find it now and the search function was no help or yielded too many choices, so I am creating a new post here.
  • Trivially, if $abc=0$, $a + b + c = 0$, and $a^2 + b^2 + c^2 = 1$, then equality holds: $a^{2m} + b^{2m} + c^{2m}= 2^{1-m},\forall m>0$.
  • I am not 100 percent sure of the inequality in the problem above, but it seems so, at least from trials.

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