The setup is as follows:
Suppose that $m$ is a given natural number. What is the greatest natural number $k$ such that for all real numbers $a,b>0$, we have $$\sqrt[k]{\frac{a^k+b^k}2}\le\frac{a^{m+1}+b^{m+1}}{a^m+b^m}$$?
Remark: Note that $$\sqrt[k]{\frac{a^k+b^k}2}\le\sqrt[l]{\frac{a^l+b^l}2}$$ when $k\le l$ by the Power mean inequality.
My attempt: I will show that $k\geq 1$, no matter what $m$ is:
Note that $$(a-b)(a^m-b^m)\geq 0$$ for all $a,b>0$ which implies that $$ab^m+a^mb \le a^{m+1}+b^{m+1}$$
and thus $$a^{m+1}+ab^m+a^mb+b^{m+1}\le2(a^{m+1}+b^{m+1})$$
i.e.
$$(a^m+b^m)(a+b)\le2(a^{m+1}+b^{m+1})$$
which is exactly the above inequality for $k=1$.
The hint.
Let $a\geq b$ and $a=xb$.
Thus, $x\geq1$ and we need $f(x)\geq0,$ where $$f(x)=\ln\left(x^{m+1}+1\right)-\ln\left(x^m+1\right)-\frac{1}{k}\ln\left(x^k+1\right)+\frac{\ln2}{k}.$$ Now, $f(1)=f'(1)=0$ and $f''(1)=\frac{2m+1-k}{4},$ which says that for $k>2m+1$ our inequality is wrong.
Thus, it's enough to prove that $$\frac{x^{m+1}+1}{x^m+1}\geq\sqrt[2m+1]{\frac{x^{2m+1}+1}{2}},$$ which is smooth: https://artofproblemsolving.com/community/c6h1750157