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.

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 $x$.
