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}}I tried this problem in this way:-
Let the real numbers are $b_2,b_3,\cdots,b_n$. such that $(b_2,b_3,\cdots,b_n)$is a permutation of the numbers given $(a_2,a_3,\cdots,a_n)$. Hence $s=b_2+b_3+\cdots+b_n$
Now denote $a_2=b_2,\ a_3=b_3,\cdots ,\ a_n=b_n$
Hence$$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_2^{1-\frac{1}{2}} +  b_3^{1-\frac{1}{3}} +  \cdots +b_n^{1-\frac{1}{n}}$$
Now denote $a_2=b_n,\ a_3=b_2,\ a_4=b_3,\cdots ,\ a_n=b_{n-1}$
Hence $$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_n^{1-\frac{1}{2}} +  b_2^{1-\frac{1}{3}} + b_3^{1-\frac{1}{4}}+ \cdots +b_{n-1}^{1-\frac{1}{n}}$$
Now denote $a_2=b_{n-1},\ a_3=b_n,\ a_4=b_2,\cdots ,\ a_n=b_{n-2}$
Hence $$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_{n-1}^{1-\frac{1}{2}} +  b_n^{1-\frac{1}{3}} + b_2^{1-\frac{1}{4}}+ \cdots +b_{n-2}^{1-\frac{1}{n}}$$
$$\vdots$$ $$\vdots$$
Now denote $a_2=b_{3},\ a_3=b_4,\ a_4=b_5,\cdots ,\ a_{n-1}=b_n,\ a_n=b_{2}$
Hence $$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_{3}^{1-\frac{1}{2}} +  b_4^{1-\frac{1}{3}} +  \cdots +b_{n}^{1-\frac{1}{n-1}}+b_{2}^{1-\frac{1}{n}}$$
Add all these and we get:- $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k^{1-\frac{1}{i}}             \right)$$
Hence we need to prove $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k^{1-\frac{1}{i}}             \right)<(n-1)\left(s+\sqrt{s}\right)$$
Now $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k^{1-\frac{1}{i}}\right)=\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k\left(b_k^{-\frac{1}{i}}\right)\right)$$
Now let a positive real number $m$ and a positve integer $p$. then $$m^{\frac{1}{p}}\leq \frac{\frac{1}{m}+\overbrace{1+1+\cdots}^{(p-1)\ \text{times}}}{p}$$
Hence $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k\left(b_k^{-\frac{1}{i}}\right)\right)\leq\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k\left(\frac{\frac{1}{b_k}+\overbrace{1+1+\cdots+1}^{(i-1)\ \text{times}}}{i}\right)\right)= \sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n \left(\frac{1}{i}\right)   +   b_k\sum\limits_{i=2}^n\left(\frac{i-1}{i}\right)\right)=
(n-1)\sum\limits_{i=2}^n \left(\frac{1}{i}\right)    +      \sum\limits_{k=2}^n \left(b_k\sum\limits_{i=2}^n\left(1-\frac{1}{i}\right)\right)=
(n-1)\sum\limits_{i=2}^n \left(\frac{1}{i}\right)    +      (n-1)\sum\limits_{k=2}^n b_k -\sum\limits_{k=2}^n\left(b_k\sum\limits_{i=2}^n\left(\frac{1}{i}\right)\right)=(n-1)s+(n-1)\sum\limits_{i=2}^n \left(\frac{1}{i}\right)-\sum\limits_{k=2}^n\left(b_k\sum\limits_{i=2}^n\left(\frac{1}{i}\right)\right)$$
Now i am stuck. Can anyone help me? If you have any other process please mention that also. This problem was in the Old and New Inequalities Volume 2. So please try to be limited in AM-GM and Cauchy-Schwarz.
This problem was proposed by George Tsintsifas in American Mathematical Monthly. So if anyone gives the original solution to this problem i will gladly welcome that.
 A: 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.)
A: 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:
$\displaystyle \left(\sum_{k=2}^n a_k^{1-\frac{1}{k}} - s\right)^2 \leqslant 4s$
