Prove $\sum\limits_{i=1}^n\frac1{x_i}\sqrt{x_i-x_{i-1}}\le\sum\limits_{i=1}^{n^2}\frac1i-\frac12$ for integers $1=x_0\le x_1\le x_2\le\cdots\le x_n$

Question

Suppose $x_{i}\in N^{+}$, and $1=x_{0}\le x_{1}\le x_{2}\le\cdots\le x_{n}$. Show that $$\sum_{i=1}^{n}\dfrac{\sqrt{x_{i}-x_{i-1}}}{x_{i}}\le\sum_{i=1}^{n^2}\dfrac{1}{i}-\dfrac{1}{2}$$

Maybe it can be proved by using the C-S inequality. But I am unable to find a solution.

Answer 1

I think the following can help.

Let for $n>1$ we have $x_n\leq n^2$.

Thus, $$\sum_{i=1}^n\frac{\sqrt{x_i-x_{i-1}}}{x_i}\leq\sum_{i=1}^n\frac{x_i-x_{i-1}}{x_i}=\sum_{i=1}^n\sum_{j=1}^{x_i-x_{i-1}}\frac{1}{x_i}\leq$$ $$\leq\sum_{i=1}^n\sum_{j=1}^{x_i-x_{i-1}}\frac{1}{x_{i-1}+j}=\sum_{i=x_0+1}^{x_n}\frac{1}{i}\leq\sum_{i=2}^{n^2}\frac{1}{i}<\sum_{i=1}^{n^2}\frac{1}{i}-\frac{1}{2}.$$ If first $x_k>n^2$.

Thus, $$\sum_{i=k}^n\frac{\sqrt{x_i-x_{i-1}}}{x_i}<\sum_{i=k}^n\frac{1}{\sqrt{x_i}}<\sum_{i=k}^n\frac{1}{n}<1$$ and $$\sum_{i=1}^{k-1}\frac{\sqrt{x_i-x_{i-1}}}{x_i}<\sum_{i=2}^{n^2}\frac{1}{i}.$$