陈博士的二次型不等式问题

(二次型不等式)设$n$为正整数, $c_1,c_2,\cdots,c_n$是复数,满足$\sum_{j=1}^{n}c_j=0$, $x_1,x_2,\cdots,x_n$是实数.证明:
$$
\sum_{j,k=1}^n{c_j\overline{c_k}\left| x_j-x_k \right|}\leqslant 0.
$$

证明.利用
$$
\int_0^{+\infty}{\frac{1-\cos \left( at \right)}{t^2}dt}=\left| a \right|
$$
有
$$\begin{aligned}
&\sum_{j,k=1}^n{c_j\overline{c_k}\left| x_j-x_k \right|} =\sum_{j,k=1}^n{c_j\overline{c_k}\int_0^{+\infty}{\frac{1-\cos \left( x_j-x_k \right) t}{t^2}dt}}
\\
&=\int_0^{+\infty}{\frac{1}{t^2}\sum_{j,k=1}^n{c_j\overline{c_k}}dt}-\int_0^{+\infty}{\frac{1}{t^2}\sum_{j,k=1}^n{c_j\overline{c_k}\left( \cos x_jt\cos x_kt+\sin x_jt\sin x_kt \right)}dt}
\\
&=\int_0^{+\infty}{\frac{1}{t^2}\left| \sum_{j=1}^n{c_j} \right|^2dt}-\int_0^{+\infty}{\frac{1}{t^2}\left( \left| \sum_{j=1}^n{c_j\cos x_j} \right|^2+\left| \sum_{j=1}^n{c_j\sin x_j} \right|^2 \right) dt}
\\
&=-\int_0^{+\infty}{\frac{1}{t^2}\left( \left| \sum_{j=1}^n{c_j\cos x_jt} \right|^2+\left| \sum_{j=1}^n{c_j\sin x_jt} \right|^2 \right) dt}\leqslant 0.
\end{aligned}$$

(二次型不等式)设$n$为正整数, $c_1,c_2,\cdots,c_n$是复数, $x_1,x_2,\cdots,x_n$为实数.证明:
$$
\sum_{j,k=1}^n{c_j\overline{c_k}\frac{1}{1+\left( x_j-x_k \right) ^2}}\geqslant 0.
$$

证明.利用
$$
\int_0^{+\infty}{e^{-t}\cos \left( at \right) dt}=\frac{1}{1+a^2}
$$
可得
$$\begin{aligned}
&\sum_{j,k=1}^n{c_j\overline{c_k}\frac{1}{1+\left( x_j-x_k \right) ^2}}=\sum_{j,k=1}^n{c_j\overline{c_k}\int_0^{+\infty}{e^{-t}\cos \left( x_j-x_k \right) tdt}}
\\
&=\int_0^{+\infty}{e^{-t}\sum_{j,k=1}^n{c_j\overline{c_k}\left( \cos x_jt\cos x_kt+\sin x_jt\sin x_kt \right)}dt}
\\
&=\int_0^{+\infty}{e^{-t}\left( \left| \sum_{j=1}^n{c_j\cos x_jt} \right|^2+\left| \sum_{j=1}^n{c_j\sin x_jt} \right|^2 \right) dt}\geqslant 0.
\end{aligned}$$

标签：infty,cos,right,不等式,二次,int,sum,陈博士,left
来源： https://www.cnblogs.com/Eufisky/p/16387947.html