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?

Fact 1: Let $n \ge 2$. Let $x_1, x_2, \cdots, x_n \ge 0$ and $x_1 > 0$. Then
$$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
 x_j} \ge \frac{2n - 1}{n^2}(x_1 + x_2 + \cdots + x_n).$$
(The proof is given at the end.)
By Fact 1, we are done.

Proof of Fact 1:
We use Mathematical Induction.
It is easy to verify the case $n = 2$.
Assume that it is true for $n$.
For $n+1$, by the inductive hypothesis, we have
\begin{align*}
 \sum_{i=1}^{n+1} \frac{x_i^2}{\sum_{j=1}^i
  x_j} 
 &= \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{2n - 1}{n^2}(x_1 + x_2 + \cdots + x_n) 
 + \frac{x_{n+1}^2}{\sum_{j=1}^{n+1} x_j}.
\end{align*}
It suffices to prove that
$$\frac{2n - 1}{n^2}(x_1 + x_2 + \cdots + x_n) 
+ \frac{x_{n+1}^2}{\sum_{j=1}^{n+1} x_j} \ge \frac{2n+1}{(n+1)^2}(x_1 + x_2 + \cdots + x_{n+1}).$$
Since the inequality is homogeneous, assume that $x_1 + x_2 + \cdots + x_n = 1$.
Letting $y = x_{n+1}$, it suffices to prove that
$$\frac{2n - 1}{n^2} 
+ \frac{y^2}{1+y} \ge \frac{2n+1}{(n+1)^2}(1 + y)$$
or
$$n^4y^2 + (-2n^3 + n^2 - 1)y + 2n^2 - 1 \ge 0$$
which is true (easy).
We are done.
