Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$ 
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$  and $k \le 2m+1$. 

My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}.$$ I need prove that for all real $x \ge 1$ and for all positive integers $m$ the inequality $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$$ holds.

Incomplete answer to date
Let $$f(x,k)=\frac{x^{m+1}+1}{x^m+1}-\left(\frac{x^k+1}2\right)^{\frac1k}\tag1$$ so that $$f_x(x,k)=\frac{\partial f(x,k)}{\partial x}=\frac{x^{m-1}(x^{m+1}+(m+1)x-m)}{(x^m+1)^2}-\frac{x^{k-1}}{x^k+1}\left(\frac{x^k+1}2\right)^{\frac1k}\tag2.$$ Clearly, $$f_x(1,k)=\frac2{2^2}-\frac12=0\tag3$$ and since \begin{align}\small\!\!\!\!\!\!\!\!\!\!\!\!f_{xx}(x,k)=\frac{\partial^2f(x,k)}{\partial x^2}&=\small\frac{mx^{m-2}(x^m+1)^2(2x^{m+1}+m(m+1)x-m(m-1))}{(x^m+1)^4}\\&\small \,\,\,\,\,\,\,\,-\frac{2mx^{2m-2}(x^{m+1}+(m+1)x-m)}{(x^m+1)^4}-\frac{(k-1)x^{k-2}}{(x^k+1)^2}\left(\frac{x^k+1}2\right)^{\frac1k}\tag4,\end{align} we have that $$f_{xx}(1,k)=\frac{m\cdot2^2(2+2m)-2m\cdot2}{2^4}-\frac{k-1}{2^2}=\frac{m(1+2m)-k+1}{2^2}>0\tag5$$ for $m\ge1$ due to the fact that $k<2m+2$. Therefore, $x=1$ is a minimum with $$f(1,k)=\frac22-1=0\tag6.$$ It remains to show that $f(x,k)>0$ for $x>1$. However, it suffices to prove that $$f_m(x)=\frac{x^{m+1}+1}{x^m+1}-\left(\frac{x^{2m+1}+1}2\right)^{\frac1{2m+1}}>0\tag7$$ since $$\left(\frac{x^k+1}2\right)^{\frac1k}\le\left(\frac{x^{2m+1}+1}2\right)^{\frac1{2m+1}}\tag8$$ from your attempt. Now $(7)$ is implied by $$2(x^{m+1}+1)^{2m+1}>(x^m+1)^{2m+1}(x^{2m+1}+1)\tag9$$ which is equivalent to $$2\sum_{n=0}^{2m+1}\binom{2m+1}nx^{nm+n}>(x^{2m+1}+1)\sum_{n=0}^{2m+1}\binom{2m+1}nx^{nm}\tag{10}.$$ Collecting the summation terms yields $$\sum_{n=0}^{2m+1}\left[\binom{2m+1}nx^{nm}(2x^n-x^{2m+1}-1)\right]>0\tag{11}$$ which is almost certainly the case.
