幸神的问题：设$\displaystyle p(x)=\sum_{i=0}^{n}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}$,若$\displaystyle a_{0}+\sum_{a_{i}<0}\left(1-\frac{i}{n}\right)C_{n}^{i}a_{i}>0$, 且$\displaystyle a_{n}+\sum_{a_{i}<0}C_{n}^{i}\cdot\frac{i}{n}\cdot a_{i}>0 $,求证：$\forall x\in[0,1]$,有$p(x)>0$.

证明：由Weight-AM-GM,有
\[ \left(1-\frac{i}{n}\right)(1-x)^{n}+\frac{i}{n}x^{n}\geq (1-x)^{n-i}x^{i} \]
这时对$p(x)$,有
\begin{align*}
 p(x)&=\sum_{i=0}^{n}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}\\
&=a_{0}(1-x)^{n}+a_{n}x^{n}+\sum_{i=1}^{n-1}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}\\
&\geq a_{0}(1-x)^{n}+a_{n}x^{n}+\sum_{a_{i}<0}C_{n}^{i}a_{i}x^{i}(1-x)^{n-i}\\
&\geq a_{0}(1-x)^{n}+ a_{n}x^{n}+\sum_{a_{i}<0}C_{n}^{i}a_{i}\left[\left(1-\frac{i}{n}\right)(1-x)^{n}+\frac{i}{n}x^{n} \right]\\
&=(1-x)^{n}\left[a_{0}+\sum_{a_{i}<0}\left(1-\frac{i}{n}\right)C_{n}^{i}a_{i}\right]+x^{n}\left[a_{n}+\sum_{a_{i}<0}C_{n}^{i}\cdot\frac{i}{n}\cdot a_{i}\right]\\
&>0
\end{align*}