Answer

问题及解答

证明不等式 $x\cdot 2^{1+\frac{1}{x}}\geqslant(x+1)^{1+\frac{1}{x}}$, 这里 $x\in(0,1]$.

Posted by haifeng on 2019-04-14 15:06:23 last update 2019-04-14 15:06:23 | Edit | Answers (1)

证明不等式

\[x\cdot 2^{1+\frac{1}{x}}\geqslant(x+1)^{1+\frac{1}{x}},\]

这里 $x\in(0,1]$.

1

Posted by haifeng on 2019-04-14 15:08:33

Pf.  等价于证明

\[
x\geqslant\Bigl(\frac{x+1}{2}\Bigr)^{1+\frac{1}{x}}
\]

两边取 $\ln$ , 得

\[
\ln x\geqslant(1+\frac{1}{x})\ln\frac{x+1}{2}.
\]

令 $\varphi(x)=\ln x-(1+\frac{1}{x})\ln\frac{x+1}{2}$.

\[
\begin{split}
\varphi'(x)&=\frac{1}{x}-\biggl[-\frac{1}{x^2}\ln\frac{x+1}{2}+(1+\frac{1}{x})\frac{1}{x+1}\biggr]\\
&=\frac{1}{x}-\biggl[-\frac{1}{x^2}\ln\frac{x+1}{2}+\frac{1}{x}\biggr]\\
&=\frac{1}{x^2}\ln\frac{x+1}{2},
\end{split}
\]

由于 $x\in(0,1]$, 故 $\varphi'(x)\leqslant 0$. 而 $\varphi(1)=0$, 因此 $\varphi(x)\geqslant 0$.