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.
 A: Since by PM $$\left(\frac{x^n+1}{2}\right)^k\geq\left(\frac{x^k+1}{2}\right)^n$$ is true for all $x\geq0$ and $n\geq k>0,$ it's enough to prove that
$$\frac{x^{m+1}+1}{x^m+1}\geq\sqrt[2m+1]{\frac{x^{2m+1}+1}{2}}$$ or $f(x)\geq0,$ where
$$f(x)=\ln\left(x^{m+1}+1\right)-\ln\left(x^m+1\right)-\frac{1}{2m+1}\ln\left(x^{2m+1}+1\right)+\frac{\ln2}{2m+1}.$$
Indeed, $$f'(x)=\frac{(m+1)x^m}{x^{m+1}+1}-\frac{mx^{m-1}}{x^m+1}-\frac{x^{2m}}{x^{2m+1}+1}=$$
$$=\frac{x^{m-1}\left(mx^{2m+2}-(m+1)x^{2m+1}+(m+1)x-m\right)}{\left(x^m+1\right)\left(x^{m+1}+1\right)\left(x^{2m+1}+1\right)}.$$
Let $g(x)=mx^{2m+2}-(m+1)x^{2m+1}+(m+1)x-m.$
Thus, $$g'(x)=m(2m+2)x^{2m+1}-(m+1)(2m+1)x^{2m}+m+1;$$
$$g''(x)=2m(m+1)(2m+1)x^{2m}-2m(m+1)(2m+1)x^{2m-1}=$$
$$=2m(m+1)(2m+1)x^{2m-1}(x-1)\geq0,$$
which says $$g'(x)\geq g'(1)=0,$$ which says
$$g(x)\geq g(1)=0,$$ which gives
$$f(x)\geq f(1)=0$$ and we are done!
By the way, we see that your inequality is true for all reals $x\geq0$ and $m>0$.
A: The result can be proved using AM $\geq$ GM.
\begin{align}
\sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}&=\sqrt[2m+1]{\prod_{i=0}^{2m} \frac{1+x^{i+1}}{1+x^{i}}}\leq\frac{1}{2m+1} \sum_{i=0}^{2m} \frac{1+x^{i+1}}{1+x^{i}} \\
&=\frac{1}{2m+1} \left[\sum_{i=0}^{m-1} \left(\frac{1+x^{i+1}}{1+x^{i}}+\frac{1+x^{2m-i+1}}{1+x^{2m-i}} \right)+\frac{1+x^{m+1}}{1+x^m}\right]\\
&=\frac{1}{2m+1}\left[ \sum_{i=0}^{m-1} \left( 2x- \frac{x-1}{1+x^{i}} -\frac{x-1}{1+x^{2m-i}}\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\
&=\frac{1}{2m+1}\left[ \sum_{i=0}^{m-1}\left( 2x- \frac{x-1}{1+x^m}\left[\frac{1+x^m}{1+x^{i}}+\frac{1+x^m}{1+x^{2m-i}}\right]\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\
&=\frac{1}{2m+1}\left[ \sum_{i=0}^{m-1}\left( 2x- \frac{x-1}{1+x^m}\left[2+\frac{x^i(x^m-1)(x^{m-i}-1)^2}{(1+x^i)(1+x^{2m-i})}\right]\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\
&\leq \frac{1}{2m+1}\left[ \sum_{i=0}^{m-1}\left(2x-(x-1)\frac{2}{1+x^{m}}\right)+ \frac{1+x^{m+1}}{1+x^{m}}\right]\\
&=\frac{1+x^{m+1}}{1+x^{m}}.
\end{align}
A: 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.
A: By the power mean inequality, with $x\geq 0, n \geq k$, we have
$$ \left( \frac{ x^n + 1 } {2} \right) ^\frac{1}{n} \geq \left( \frac{x^k + 1 } { 2} \right) ^ \frac{1}{k}. $$
Hence, all that we need to show is:
$$ \frac{ x^{m+1} + 1 } { x^m + 1} \geq \left( \frac{ x^{2m+1 } + 1 } {x^0 + 1 } \right)  ^ \frac{1}{2m+1} \quad (1) . $$

We will show that
$$ \left(\frac{1+x^{m+1}}{1+x^m}\right)^2 \geq \frac{1+x^{m+1+i}}{1+x^{m+i}} \times \frac{1+x^{m+1-i}}{1+x^{m-i}} \quad (2) $$
By cross multiplying and expanding terms (Thanks Wolfram), this is equivalent to:
$$ x^{m-i} (x^i-1)^2(x-1)(x^{2m+1} - 1) \geq 0. $$
This is obviously true for $ x \geq 0$.
(Easier to see by splitting up into cases $x\geq 1, x < 1$, which might explain where that condition came from.)

Then, (1) follows by multiplying the chain of inequalities of (2), going from $i=0$ to $m$.

Note: The (pseudo-)reasoning behind why (2) should be true, is that when we write it as $\prod ( 1+ x^{a_i}) \geq \prod ( 1+ x^{b_i})$, with $a_i, b_i$ in decreasing order, we have $ \sum a_i = \sum b_i$. Since $a_1 = m+i < m+i+1 = b_1$, it follows that after cancelling the $x^{\sum a_i}$ term on both sides, the LHS has a larger exponent, so it will be true for large enough $x$.
A: for all $m\in \mathbb{N}$ and $x\geq 1$ we have $x^{2m+1}\geq 1$ then $2x^{2m+1}\geq x^{2m+1}+1$
Notice that $\frac{x^{m+1}+1}{x^m+1} = x - \frac{x-1}{x^m+1} \geq x$ because $x\geq 1$
Then $$2\left(\frac{x^{m+1}+1}{x^m+1}\right)^{2m+1}\geq 2x^{2m+1}\geq x^{2m+1}+1$$
We obtain easily that $$\frac{x^{m+1}+1}{x^m+1} \geq \sqrt[2m+1]{\frac{x^{2m+1}+1}{2}}$$
