<s>[INST] How prove $\sum_{i=2}^na_i^{1-\frac{1}{i}} < S+2\sqrt{S}$ for $S=a_2+\dots +a_n.$ Let $a_2,\dots,a_n>0$ and $S=a_2+\dots +a_n.$ How prove $\sum_{i=2}^na_i^{1-\frac{1}{i}} < S+2\sqrt{S}.$
 [/INST] Because $\sum_{i=1}^na_i^{1-\frac{1}{i}}=\sum_{i=2}^na_i^{1-\frac{1}{i}}+1 < S+2\sqrt{S}+1=(\sqrt S +1)^2 \Leftrightarrow \sum_{i=2}^na_i^{1-\frac{1}{i}} < S+2\sqrt{S}.$
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It's sufficient to show $(\sqrt S +1)^2 >  {\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 $m=n$:
$(\sqrt S_{n-1} + 1)^2 +a_n^{1-\frac 1n}=S_{n-1}+1+2\sqrt S_{n-1}+a_n^{1-\frac 1n} 
\\ =S_n+1+2\sqrt S_{n-1}<S_n+1+2\sqrt S_{n}=(\sqrt S_n +1)^2$
Hence $\sum_{i=1}^{n}a_i^{1-\frac{1}{i}} < (\sqrt S_{n-1} + 1)^2+a_n^{1-\frac 1n} < (\sqrt S_n +1)^2$
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