China Mathematical Competition

When $n\ge 2$, $S_n\ge 2S_{n-1}$ is equivalent to $$x_n\ge x_1+\cdots +x_{n-1}.①$$ Let $C=\frac{1}{4}{x}_1$. We will prove $$x_n\ge C\cdot 2^n,n=1,2,\dots \stackrel{◯}{2}$$ by induction.

When $n=1$, it is obviously true. When $n=2$, we have $x_2\ge x_1=C\cdot 2^2$.

When $n\ge 3$, assume $x_k\ge C\cdot 2^k$, $k=1,2,\dots ,n-1$. Then from ①, we have $$\begin{array}{rl}{x}_n& \ge x_1+(x_2+\cdots +x_{n-1})\\ & \ge x_1+(C\cdot 2^2+\cdots +C\cdot 2^{n-1})\\ & =C(2^2+2^2+2^3+\cdots +2^{n-1})=C\cdot 2^n.\end{array}$$ Therefore, ② holds for every $n$.