Let be $a,b,c$ real numbers from $\left[ -1,1 \right]$ satisfying :
$$1+2abc\ge a^{2}+b^{2}+c^{2}$$
Show that :
$$\forall n\in \mathbb{N} \ \ \ \ 1+2\left( abc \right)^{n}\ge a^{2n}+b^{2n}+c^{2n}$$
I tried using induction but I can't find way to lower the term $a^{2n+2}+b^{2n+2}+c^{2n+2}$ properly. Any suggestions or hints
This problem was on IMC 2010.
Let $ab\geq0.$
Thus, since the condition gives $$(ab-c)^2\leq(1-a^2)(1-b^2)$$ and we need to prove that $$(a^nb^n-c^n)^2\leq(1-a^{2n})(1-b^{2n}),$$ by C-S we obtain: $$(a^nb^n-c^n)^2=(ab-c)^2\left(\sum_{k-0}^{n-1}(ab)^{n-1-k}c^k\right)^2\leq$$ $$\leq(ab-c)^2\left(\sum_{k-0}^{n-1}(ab)^{k}\right)^2\leq(ab-c)^2\sum_{k=0}^{n-1}a^{2k}\sum_{k=0}^{n-1}b^{2k}\leq$$ $$\leq(1-a^2)(1-b^2)\sum_{k=0}^{n-1}a^{2k}\sum_{k=0}^{n-1}b^{2k}=(1-a^{2n})(1-b^{2n}).$$
