Let $a,b,c\in[-1,1]$ be such that $$2abc+1\geq a^2+b^2+c^2.$$ Prove that $$2a^nb^nc^n+1\geq a^{2n}+b^{2n}+c^{2n}$$ for any positive integer $n$.
The case $n=1$ is of course the same as the assumption. For $n=2$, squaring the assumption gives $$4a^2b^2c^2+4abc+1\geq a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$$ which is quite different than what is desired.
Here is a partial solution : $a>0,b>0,c>0$ case
$b=\cos{B},c=\cos{C}, 0 \le B,C \le \dfrac{\pi}{2}$
$a^2-2abc\le 1-b^2-c^2 \iff bc-\sqrt{(1-b^2)(1-c^2)} \le a \le bc+\sqrt{(1-b^2)(1-c^2)} \iff \cos{(B+C)} \le a \le \cos{(B-C)}$
now we need to prove:
$b^{n}c^{n}-\sqrt{(1-b^{2n})(1-c^{2n})} \le a^{n} \le b^nc^n+\sqrt{(1-b^{2n})(1-c^{2n})} \\$
upbound $\implies \cos^n{(B-C)} \le \cos^n{B}\cos^n{C}+\sqrt{(1-\cos^{2n}{B})(1-\cos^{2n}{C})} \\ \iff \cos^n{(B-C)}-(\cos{B}\cos{C})^n\le \sqrt{(1-\cos^2{B})(1+\cos^2 B+(\cos^2B)^2+...+(\cos^2B)^{n-1})(1-\cos^2{C})(1+\cos^2 C+(\cos^2C)^2+...+(\cos^2C)^{n-1})} \\ \iff (\cos {(B-C)}-\cos B \cos C)(\cos^{n-1} (B-C) +\cos^{n-2}(B-C) (\cos B \cos C)+\cos^{n-3}(B-C) (\cos B \cos C)^2+...+\cos (B-C) (\cos B \cos C)^{n-2}+ (\cos B \cos C)^{n-1}\le \sin B \sin C\sqrt{(1+b^2+(b^2)^2+...(b^2)^{n-1})(1+c^2+(c^2)^2+...+((c^2)^{n-1})} $
$\cos (B-C) -\cos B \cos C= \sin B \sin C \ge 0$ , it remains
$\cos^{n-1} (B-C) + \cos^{n-2} (B-C) (bc)+ \cos^{n-3} (B-C) (bc)^2+...\cos (B-C) (bc)^{n-2}+(bc)^{n-1} \le \sqrt{(1+b^2+(b^2)^2+...(b^2)^{n-1})(1+c^2+(c^2)^2+...+((c^2)^{n-1})} $
LHS $\le 1+bc+(bc)^2+...(bc)^{n-1}$ because $0 \le (\cos (B-C))^k \le 1$
RHS $\ge 1+bc+(bc)^2+...(bc)^{n-1}$ it is Cauchy.
for low bound, it is similar.
for $abc \le 0$ , it is easy to prove as $2(abc)^n+1 \ge 2abc+1 \ge a^2+b^2+c^2 \ge a^{2n}+b^{2n}+c^{2n}$
