# Inequality relating harmonic numbers

I'm stuck at proving the following inequality: For any m,n positive integers with $m \leq2^n$ we have $$\sqrt{\frac{2^n}{1}} + \sqrt{\frac{2^n}{2}}+...+\sqrt{\frac{2^n}{m-1}}+\sqrt{\frac{2^n}{m}}\leq \sqrt{m2^nH_m}$$ where $H_m$ is the m-th harmonic number... The $\sqrt{2^n}$ factors out so I dont even understand why it would be there, and subadditivity of square root takes me in the other direction.

Any hint would be appreciated ! Thanks

You are certainly right that the $2^n$ make no difference. The rest follows from the Schwarz Inequality.
In the notation used in the linked article, we let $x_i=\frac{1}{\sqrt{i}}$ and $y_i=1$. Then the Schwarz Inequality tells us that $$\left(\sum_{i=1}^m \frac{1}{\sqrt{i}}\right)^2 \le \left(\sum_{i=1}^m \frac{1}{i}\right)\left(\sum_{i=1}^m 1\right).$$ Taking square roots yields the desired result.