Ugly solution: We want $ \frac {1}{(1-x)^n }+\frac {1}{(1+x)^n }\ge(1-x)^m+(1+x)^m$ for $ |x|<1$. Now \[ LHS-RHS=2\sum_{k=0}^{\infty}\left[ \binom{-n}{2k}-\binom{m}{2k}\right]x^{2k}=2\sum_{k=1}^{\infty}\left[ \binom{n+2k-1}{2k}-\binom{m}{2k}\right]x^{2k}\] from which it follows immediately that $ m\le n+1$.

WLOG we may assume $a>b\Rightarrow a>1>b$, so $a^m+b^m\ge a^{m-1}+b^{m-1}\Longleftrightarrow (a-1)a^{m-1}\ge (1-b)b^{m-1}\Longleftrightarrow a^{m-1}\ge b^{m-1}$ which is true. So, by induction, result that $a^m+b^m\ge a^{m-1}+b^{m-1}\ge\ldots\ge a+b$. On the other hand $ \frac{1}{a^n}+\frac{1}{b^n}\ge a^n+b^n \Longleftrightarrow (a^n+b^n)(1-a^nb^n)\ge 0$ which is true, so $m\ge n$. Suppose $m>n$, we have $ \frac{1}{a^n}+\frac{1}{b^n}\ge a^m+b^m \Longleftrightarrow \frac{a^{m+n}-1}{a^n}\le \frac{1-b^{m+n}}{b^n}\Longleftrightarrow$ $a^{m-1}+a^{m-2}+\ldots+a+1+\frac{1}{a}+\ldots+\frac{1}{a^n}\le b^{m-1}+b^{m-2}+\ldots+b+1+\frac{1}{b}+\ldots+\frac{1}{b^n}\Longleftrightarrow$ $(a^{m-1}-b^{m-1})+\ldots+(a^{n+1}-b^{n+1})\le \frac{(a-b)(1-ab)}{ab}+\ldots+\frac{(a^n-b^n)(1-a^nb^n)}{a^nb^n}$ which is true if $m=n+1$ because $LHS=0$ and $RHS\ge 0$. If $m>n+1$, after dividing with $a-b$, and passing to the limit when $a \searrow 1( b\nearrow 1, ab\rightarrow 1)$ we get $LHS\rightarrow (m-1)+\ldots +(n+1)$ and $RHS\rightarrow 0$ which is impossible, so $m\le n+1$.