Let $a_2,a_3,\cdots,a_n$ be positive real numbers and $s=a_2+a_3+\cdots+a_n$. Show that $\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}<s+2\sqrt{s}$

We can prove that $a_k^{1-1/k} < a_k + \frac{2}{k} \sqrt{a_k}$. Indeed, if $a_k \ge 1$, it is obvious; and if $0 < a_k < 1$, by Bernoulli inequality $(1+x)^r \le 1 + rx$ for $0 < r \le 1$ and $x > -1$, we have $a_k^{1-1/k} = a_k (a_k^{-1/2})^{2/k} = a_k(1 + a_k^{-1/2} - 1)^{2/k} \le a_k [1 + (a_k^{-1/2} - 1)\frac{2}{k}] < a_k + \frac{2}{k}\sqrt{a_k}$.

Thus, by Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align} \sum_{k=2}^n a_k^{1-1/k} &< \sum_{k=2}^n a_k + \sum_{k=2}^n \frac{2}{k} \sqrt{a_k}\\ &= \sum_{k=2}^n a_k + \sqrt{\sum_{k=2}^n \frac{4}{k^2}}\sqrt{\sum_{k=2}^n a_k}\\ &= s + 2\sqrt{\sum_{k=1}^n \frac{1}{k^2} - 1}\ \sqrt{s}\\ &< s + 2\sqrt{s} \end{align} where we have used $\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$ to get $\sqrt{\sum_{k=1}^n \frac{1}{k^2} - 1} < \sqrt{\frac{\pi^2}{6} - 1} < 1$. (Q. E. D.)

Inspired by the solution published in the American Mathematical Monthly:

We can suppose that $\ \forall k \ , \ 0<a_k<1 $.

$\displaystyle \sum_{k=2}^n a_k^{1-\frac{1}{k}} - s = \sum_{k=2}^n \left( a_k^{1-\frac{1}{k}}-a_k\right) = \sum_{k=2}^n a_k^{\frac{1}{2}}\left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right)$

By using the Cauchy-Schwarz inequality:
$\displaystyle \left(\sum_{k=2}^n a_k^{1-\frac{1}{k}} - s\right)^2 \leqslant \sum_{k=2}^n a_k \ \sum_{k=2}^n \left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right)^2 = s\ \sum_{k=2}^n \left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right)^2 $

We have, for all $k\geqslant 3$:
$\left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right)^2= a_k^{1-\frac{2}{k}}\left( 1-\exp\left( \dfrac{\ln a_k}{k}\right)\right)^2 \leqslant a_k^{\frac{1}{3}} \ln^2(a_k)\dfrac{1}{k^2} \leq \dfrac{36}{k^2e^2}$

So:
$\displaystyle \sum_{k=2}^n \left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right)^2 \leqslant 1+\dfrac{36}{e^2}\sum_{k=3}^n\dfrac{1}{k^2} \leqslant 1+\dfrac{36}{e^2}\left(\dfrac{\pi^2}{6}-\dfrac{5}{4}\right) < 4$

Eventually:
$\displaystyle \left(\sum_{k=2}^n a_k^{1-\frac{1}{k}} - s\right)^2 \leqslant 4s$