Prove that for every $k, m \in \mathbb N$: $\sqrt[k+m]{k^m\cdot m^k} \le \frac{k+m}{2}.$ Prove that for every $k, m \in \mathbb N$: $$\sqrt[k+m]{k^m\cdot m^k} \le \frac{k+m}{2}.$$
My idea:
$$\sqrt[k+m]{\left(\frac 1k\right)^m \left(\frac 1m\right)^k} \le \frac{\frac mk + \frac km}{k+m}$$
If I could prove that $\tfrac mk + \frac km\le 2$ I would get a thesis, but I have no idea how to do it.
 A: By weighted AM-GM, we have
$$\sqrt[k+m]{m^k.k^m}\le \frac{km+mk}{k+m}$$
By plain HM-AM, we have
$$\frac{2km}{k+m} \le \frac{k+m}{2}$$
Combining these two we prove the result.
A: Notice $m^2+k^2 < (m+k)^2$. Hence
$$\frac{\frac{m^2+k^2}{km}}{k+m} < \frac{m+k}{mk}=\frac{1}{k}+\frac{1}{m} \leq 2$$
A: Consider $m,\cdots,m, k\cdots, k$ where the number of $k$-s equals $m$ and vice versa.
Applying GM-AM we find $\sqrt[k+m]{k^m \cdot m^k}\le {{2km}\over {m+k}}$.
And this is $\le {{k+m}\over 2}$ since cross-multiplication yields $4km\le (m+k)^2\iff (m-k)^2\ge 0$ which is true.
The last inequality is also GM-HM.
A: Another way that I attempted.
I prove the same when $k,n>0$ are real. Taking logs, it suffices to show
$log\left(\frac{k+m}{2}\right)\ge \frac{m}{k+m}\log k + \frac{k}{m+k}\log m$.
As $x\mapsto \log x$ is concave, Jensen's inequality tells us
$\frac{m}{k+m}\log k + \frac{k}{m+k}\log m\le \log\left(\frac{2km}{k+m}\right)$.
Finally, we conclude by recalling
$\frac{k+m}{2}\ge \frac{2km}{k+m}\iff (k-m)^2\ge 0$ and the fact $\log$ is increasing.
