<s>[INST] prove $\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n}-\frac{1}{n^2}$ 
Let $x_1,x_2,\cdots,x_n$ be $n$ non-negative numbers such that
$x_1+x_2+\cdots+x_n=1$. Show $$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
 x_j}\ge \frac{2}{n}-\frac{1}{n^2}.$$

We may consider using Cauchy. That is
$$\sum_{i=1}^n\sum_{j=1}^i x_j\cdot \sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
 x_j}\ge \left(\sum_{i=1}^nx_i\right)^2=1.$$
How to go on?
 [/INST] Base Case:- $n=1$
$$\sum_{i=1}^1 \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{1}-\frac{1}{1^2}$$
$$1 \ge 1$$
Thus the base case holds.
Hypothesis:-
$$\sum_{i=1}^{n+1} \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n+1}-\frac{1}{{(n+1)}^2}$$
And now for the step:-
$$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j} + \frac{x_{n+1}^2}{\sum_{j=1}^{n+1} x_j}\ge \frac{2}{n}-\frac{1}{n^2} + \frac{x_{n+1}^2}{\sum_{j=1}^{n+1} x_j}$$
$$\sum_{i=1}^{n+1} \frac{x_i^2}{\sum_{j=1}^i x_j} \ge \frac{2}{n}-\frac{1}{n^2} + \frac{x_{n+1}^2}{1}$$
This is as $x_1+x_2+\cdots+x_n+x_{n+1}=1$
As $n \ge 1$ so:-
$$2n^2\ge1$$
$$\Rightarrow 2n(n+1)\ge2n+1$$
$$\Rightarrow 2 \ge \frac{2n+1}{n(n+1)}$$
$$\Rightarrow 2\cdot\left[\frac{1}{n(n+1)}\right] \ge \frac{2n+1}{n^2(n+1)^2}$$
$$\Rightarrow 2\cdot\left[\frac{1}{n} - \frac{1}{n+1}\right]\ge \frac{1}{n^2}-\frac{1}{(n+1)^2}$$
$$\Rightarrow\frac{2}{n}-\frac{1}{n^2}\ge \frac{2}{n+1} -\frac{2}{(n+1)^2} $$
$$\Rightarrow\frac{2}{n}-\frac{1}{n^2} + x_{n+1}^2\ge \frac{2}{n+1} -\frac{1}{(n+1)^2} $$
This is as $x_{n+1}$ is non-negative.
Thus we finally get:-
$$\sum_{i=1}^{n+1} \frac{x_i^2}{\sum_{j=1}^i x_j} \ge \frac{2}{n}-\frac{1}{n^2} + x_{n+1}^2\ge \frac{2}{n+1} -\frac{1}{(n+1)^2} $$
completing the step.
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