Problem 1568

Prove that when $n=1$, $x_{n}=1$, the proposition holds. Now assume $n>1$, and let $x_{n}=x$. Clearly, $x \neq 1$, and we have $x^{n}=\frac{x^{n}-1}{x-1}$, which means
$$x^{n+1}-2 x^{n}+1=0$$

From (1), we get $x^{n}(2-x)=1$, so $\text{x1-x n -x (1-x n ) = (1-x n )(1-x)>0}$$$$$

This is a contradiction, so $x \geqslant 1$.
Next, $x \geqslant 2-\frac{2}{n+1}$. In fact, if $x\frac{1}{n}\left(2-x+\frac{x-1}{n}\right)$ (this is because when the sum of two numbers is a constant $2-x+\frac{x}{n}$, the smaller the smaller number, the smaller the product), i.e.,
$$x(2-x)>2-x+\frac{x-1}{n}$$

Repeating this process, we get
$$x^{n}(2-x)>x^{n-1}\left(2-x+\frac{x-1}{n}\right)>\cdots>2-x+\frac{x-1}{n} \cdot n=1$$

This contradicts (1), which proves that $x \geqslant 2-\frac{2}{n+1}$.
For $y>x \geqslant 2-\frac{2}{n+1}$, we have $\frac{y}{n}>\frac{x}{n} \geqslant 2-x$, so
$$\frac{x}{n}(2-x)>\frac{y}{n}\left(2-x+\frac{y-x}{n}\right)$$

Therefore,
$$x^{n}(2-x)>x^{n-1} y\left(2-x+\frac{y-x}{n}\right)>\cdots>y^{n}\left(2-x+\frac{y-x}{n} \cdot n\right)=y^{n}(2-y)$$

Thus, if $x\left(2-\frac{1}{2 n-1}\right)^{n} \cdot \frac{1}{2^{n-1}}=2\left(1-\frac{1}{2^{n}}\right)^{n}>1$$

This contradicts (1), so $x \geqslant 2-\frac{1}{2^{n}}$.
This proves the required inequality.