How prove $\sum_{i=2}^na_i^{1-\frac{1}{i}} < S+2\sqrt{S}$ for $S=a_2+\dots +a_n.$ Let $a_2,\dots,a_n&gt;0$ and $S=a_2+\dots +a_n.$ How prove $\sum_{i=2}^na_i^{1-\frac{1}{i}} &lt; S+2\sqrt{S}.$
Because $\sum_{i=1}^na_i^{1-\frac{1}{i}}=\sum_{i=2}^na_i^{1-\frac{1}{i}}+1 &lt; S+2\sqrt{S}+1=(\sqrt S +1)^2 \Leftrightarrow \sum_{i=2}^na_i^{1-\frac{1}{i}} &lt; S+2\sqrt{S}.$

$ \ $

It's sufficient to show $(\sqrt S +1)^2 &gt;  {\sum_{i=1}^na_i^{1-\frac{1}{i}}}\tag 1 $

Suppose $S_n=a_2+\dots +a_n.$ 

Prove by induction:

When $n=1,2$ (1) holds obviously.

If $m=n-1\ge 2$ holds,then for $m=n$:

$(\sqrt S_{n-1} + 1)^2 +a_n^{1-\frac 1n}=S_{n-1}+1+2\sqrt S_{n-1}+a_n^{1-\frac 1n} 
\\ =S_n+1+2\sqrt S_{n-1}&lt;S_n+1+2\sqrt S_{n}=(\sqrt S_n +1)^2$

Hence $\sum_{i=1}^{n}a_i^{1-\frac{1}{i}} &lt; (\sqrt S_{n-1} + 1)^2+a_n^{1-\frac 1n} &lt; (\sqrt S_n +1)^2$
Because $\sum_{i=1}^na_i^{1-\frac{1}{i}}=\sum_{i=2}^na_i^{1-\frac{1}{i}}+1 &lt; S+2\sqrt{S}+1=(\sqrt S +1)^2 \Leftrightarrow \sum_{i=2}^na_i^{1-\frac{1}{i}} &lt; S+2\sqrt{S}.$

$ \ $

It's sufficient to show $(\sqrt S +1)^2 &gt;  {\sum_{i=1}^na_i^{1-\frac{1}{i}}}\tag 1 $

Suppose $S_n=a_2+\dots +a_n.$ 

Prove by induction:

When $n=1,2$ (1) holds obviously.

If $m=n-1\ge 2$ holds,then for $m=n$:

$(\sqrt S_{n-1} + 1)^2 +a_n^{1-\frac 1n}=S_{n-1}+1+2\sqrt S_{n-1}+a_n^{1-\frac 1n} 
\\ =S_n+1+2\sqrt S_{n-1}&lt;S_n+1+2\sqrt S_{n}=(\sqrt S_n +1)^2$

Hence $\sum_{i=1}^{n}a_i^{1-\frac{1}{i}} &lt; (\sqrt S_{n-1} + 1)^2+a_n^{1-\frac 1n} &lt; (\sqrt S_n +1)^2$