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由$\ln x$上凸，故对$x,y>0$以及$\lambda\in(0,1)$有
\[\lambda\ln x+(1-\lambda)\ln y\le\ln(\lambda x+(1-\lambda)y).\]
结合上述不等式，可得
\begin{align*}
\ln x_{n+1} &= \frac1{n+1}\sum_{k=1}^{n+1}\ln\frac{ka+(n+2-k)b}{n+2} \\
&= \frac1n\sum_{k=0}^n\left(\frac k{n+1}\ln\frac{(k+1)a+(n+1-k)b}{n+2}+\frac{n-k}{n+1}\ln\frac{(k+1)a+(n+1-k)b}{n+2}\right) \\
&= \frac1n\sum_{k=1}^n\left(\frac {n-k+1}{n+1}\ln\frac{ka+(n+2-k)b}{n+2}+\frac k{n+1}\ln\frac{(k+1)a+(n+1-k)b}{n+2}\right) \\
&\le\frac1n\sum_{k=1}^n\ln\frac{ka+(n+1-k)b}{n+1} \\
&=\ln x_n.
\end{align*}
因此得到$x_n\le x_{n+1}$对任意$n\in\mathbb{N}^+$成立. |
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睡神
Posted 2015-12-3 00:27
