$$\bex a_n\geq 0\ra \vsm{n}a_n\leq \sqrt{\pi}\sex{\vsm{n}a_n^2}^{1/4} \sex{\vsm{n}n^2a_n^2}^{1/4}, \eex$$ $$\bex \int_0^\infty |f(x)|\rd x \leq\sqrt{\pi}\sex{ \int_0^\infty f^2(x)\rd x }^{1/4}\sex{ \int_0^\infty x^2f^2(x)\rd x }^{1/4}. \eex$$
证明: 设 $$\bex \al=\vsm{n}n^2a_n^2,\quad \beta=\vsm{n}a_n^2, \eex$$ 则 $$\beex \bea \sex{\vsm{n}a_n}^2&=\sex{\vsm{n}a_n\sqrt{\al+\beta n^2}\frac{1}{\sqrt{\al+\beta n^2}}}^2 \leq \vsm{n}a_n^2(\al+\beta n^2)\vsm{n}\frac{1}{\al+\beta n^2}\\ &\leq 2\al \beta \int_0^\infty \frac{1}{\al+\beta x^2}\rd x =\pi \al\beta. \eea \eeex$$
