设$f:[0,1]\to \mathbb{R}$是一个可微函数具有连续导数，且$f(1)=0$,证明:
$$4\int_{0}^{1}x^2|f'(x)|^{2}dx\geq \int_{0}^{1}|f(x)|^2dx+\left(\int_{0}^{1}|f(x)|dx\right)^{2}$$
证明：为了方便起见，我们令
$$A=\int_{0}^{1}|f(x)|dx$$
由Cauchy-Schwarz不等式，我们有
$$LHS=4\left(\int_{0}^{1}x^{2}|f'(x)|^{2}dx\right)\left(\int_{0}^{1}(|f(x)|+A)^{2}dx\right)\geq 4\left(\int_{0}^{1}x|f'(x)|\cdot|f(x)|dx+A\int_{0}^{1}x|f'(x)|dx \right)^{2}$$
这时，注意到
$$\int_{0}^{1}x|f'(x)|\cdot|f(x)|dx\geq \left|\int_{0}^{1}xf'(x)dx\right|=\frac{1}{2}\int_{0}^{1}|f(x)|^{2}dx$$
和
$$\int_{0}^{1}|f(x)|dx=\int_{0}^{1}\left|\int_{t}^{1}f'(x)dx \right|\leq \int_{0}^{1}\int_{t}^{1}|f'(x)|dx= \int_{0}^{1}x|f'(x)|dx$$
$$LHS\geq \left(\int_{0}^{1}|f(x)|^{2}dx+2A\int_{0}^{1}|f(x)|dx\right)^{2}$$
于是，只要证明
$$\left(\int_{0}^{1}|f(x)|^{2}dx+2A\int_{0}^{1}|f(x)|dx\right)^{2}\geq \left[\int_{0}^{1}|f(x)|^2dx+\left(\int_{0}^{1}|f(x)|dx\right)^{2} \right]\left(\int_{0}^{1}(|f(x)|+A)^{2}dx\right)$$
经过简单的化简运算，就是
$$\left(\int_{0}^{1}|f(x)|dx\right)^{4}\geq 0$$
显然成立。Hence we are done!