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?

We'll assume $x_1>0,$ since the left hand side is undefined if $x_1=0.$ But if ww treat the first term as zero when $x_1=0,$ we can just use that the right side is decreasing on $n.$
Try induction on $n.$ When $n=1,$ you have $x_1=1,$ the inequality becomes $1\geq 1.$
If true for $n,$ take $x_1,\dots,x_{n+1}.$
Since $x_{1}>0,$ then the $x_{n+1}<1.$
Let $y_i=\frac{x_i}{1-x_{n+1}}$ for $i=1,\dots,n.$
Then $\sum_1^n y_i=1,$ so we have:
$$\frac2n-\frac1{n^2}\leq \sum_{i=1}^n \frac{y_i^2}{\sum_{j=1}^i y_j}=\frac1{1-x_{n+1}}\sum_{i=1}^n\frac{x_i^2}{\sum_{j=1}^i x_j}$$
So $$\begin{align}\sum_{i=1}^{n+1}\frac{x_i^2}{\sum_{j=1}^i x_j}&\geq (1-x_{n+1})\left(\frac2n-\frac1{n^2}\right)+x_{n+1}^2\\\end{align}$$
In general, the minimum value for $f(x)=(1-x)a+x^2$ when $0<a\leq 1$ and $x\in [0,1]$ is at $x=a/2$ and the value is $a-\frac{a^2}{4}=1-(1-a/2)^2.$
So this is the recursion for the absolute minimum value of the left hand side:
$$m_1=1, m_{n+1}=\inf_{m\in[m_n,1]}1-(1-m/2)^2=1-(1-m_n/2)^2.$$
Because $1-(1-u/2)^2=u-\frac{u^2}4$ is increasing on $[0,1].$
We have, if $m_n\geq \frac2n-\frac1{n^2}$ then:
$$\begin{align}m_{n+1}&\geq 1-\left(1-\frac 1n+\frac1{2n^2}\right)^2\end{align}$$
Now, $$\frac2{n+1}-\frac1{(n+1)^2}=1-\left(1-\frac1{n+1}\right)^2$$
So you need $$\frac1{n+1}\leq \frac1{n}-\frac1{2n^2}=\frac{2n-1}{2n^2}$$
or $$2n^2\leq (n+1)(2n-1)=2n^2+n-1$$ which is obviously true for $n\geq1.$
