<s>[INST] 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.
 [/INST] 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 $m>0$.
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