Prove that $\sum_{i=1}^n \sqrt{ \binom{n}{i}} \le [n(2^n -1) ]^{1/2}$ for $n \ge 2$

Question

Prove that $\sum_{i=1}^n \sqrt{ \binom{n}{i}} \le [n(2^n -1) ]^{1/2}$ for $n \ge 2$

What I have tried:

$$2^n -1= \sum_{i=1}^n \binom{n}{i}\Rightarrow \sum_{i=1}^n \sqrt{ \binom{n}{i}} \stackrel{?}{\le} \sqrt{n} \sqrt{\sum_{i=1}^n \binom{n}{i}}$$

and I tried using AM-GM inequality, but that just leaves us at $$\frac {\sum_{i=1}^n \binom{n}{i}}{n} \stackrel{?}\ge \sqrt[n]{ \prod_{i=1}^n \binom{n}{i}}.$$ How to go about it now? Please give only a hint and not a complete solution of the question.

Answer 1

By Cauchy-Schwarz inequality, $$\sum_{i=1}^n \sqrt{ \binom{n}{i}} \le \left(\sum_{i=1}^n \binom{n}{i} \right)^{1/2} \cdot\left(\sum_{i=1}^n 1^2\right)^{1/2}.$$