设$f(x)\in C$是实值函数，且满足
$$\int_{0}^{1}{f(x)dx}=\int_{0}^{1}{xf(x)dx}=\cdots=\int_{0}^{1}{x^{n-1}f(x)dx}=1$$

(tian_275461)
证明:
$$\int_{0}^{1}{(f(x))^{2}dx}\geq n^2$$
证明:
首先，我们考虑多项式$P(x)$
$$P(x)=a_{0}+a_{1}x+\cdots+a_{n-1}x^{n-1}$$
若多项式$P(x)$也满足上面的条件，那么
$$\int_{0}^{1}{(P(x))^{2}dx}=a_{0}+a_{1}+\cdots+a_{n-1}$$
为了求出系数$a_{i}$我们再次利用条件.
$$\int_{0}^{1}{x^{k}P(x)dx}=1 \qquad k=0,1,\ldots n-1$$
$$\Rightarrow \frac{a_{0}}{k+1}+\frac{a_{1}}{k+2}+\cdots+\frac{a_{n-1}}{k+n}=1 \qquad  k=0,1,\ldots n-1$$
设
$$H(x)=\frac{a_{0}}{x+1}+\frac{a_{1}}{x+2}+\cdots+\frac{a_{n-1}}{x+n}-1$$
则显然有
$$H(0)=H(1)=\cdots=H(n-1)=0$$
$H(x)$应该有
$$H(x)=\frac{A\cdot x\cdot(x-1)\cdot(x-2)\cdots(x-n+1)}{(x+1)(x+2)\cdots(x+n)}$$
通过对比系数得$A=-1$,及
$$a_{k}=(-1)^{n-k-1}\frac{(n+k)!}{(k!)^{2}\cdot(n-k-1)!}  \qquad  k=0,1,\ldots n-1$$
用数学归纳法不难证明
$$\sum_{k=0}^{n-1}{a_{k}}=n^{2}$$
所以，若多项式$P(x)$满足上面的性质，则
$$\int_{0}^{1}{(P(x))^{2}dx}=a_{0}+a_{1}+\cdots+a_{n-1}=n^2$$
取满足以上条件的多项式$P(x)$
应用Cauchy-Schwarz不等式
$$\int_{0}^{1}{(P(x))^2dx}\int_{0}^{1}{(f(x))^2dx}\geq \left(\int_{0}^{1}{P(x)f(x)dx} \right)^{2}=n^4$$
$$\Rightarrow \int_{0}^{1}{(f(x))^{2}dx}\geq n^2$$
Done!