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Author |
lemondian
Posted at 2023-10-28 11:05:42
Last edited by hbghlyj at 2025-3-22 09:04:11有人给我发来（1）的以下证明，大家看看问题出在哪？
定理 设 $\lambda \leqslant \frac{2}{3}, s>0$，则
\[
L^s<\lambda G^s+(1-\lambda) A^s
\]
引理 设 $x>0$，则 $1<\frac{\operatorname{sh} x}{x}<\operatorname{ch} x$．
证明 不妨设 $b>a$，令 $t=\ln b-\ln a$，则 $\mathrm{e}^t=\frac{b}{a}>1, t>0$，所以
\[
\begin{aligned}
& L^s-\lambda G^s-(1-\lambda) A^s \\
&= \left(\frac{b-a}{\ln b-\ln a}\right)^s-\lambda(\sqrt{a b})^s+(\lambda-1)\left(\frac{a+b}{2}\right)^s \\
&= a^s\left[\left(\frac{\frac{b}{a}-1}{\ln \frac{b}{a}}\right)^s-\lambda\left(\sqrt{\frac{b}{a}}\right)^s+(\lambda-1)\left(\frac{1+\frac{b}{a}}{2}\right)^s\right] \\
&= a^s\left[\left(\frac{\mathrm{e}^t-1}{t}\right)^s-\lambda \mathrm{e}^{\frac{t}{2}}+(\lambda-1)\left(\frac{1+\mathrm{e}^t}{2}\right)^s\right] \\
&= a^s \mathrm{e}^{\frac{s t}{2}}\left[\left(\frac{\mathrm{e}^{\frac{t}{2}}-\mathrm{e}^{-\frac{t}{2}}}{t}\right)^s-\lambda+(\lambda-1)\left(\frac{\mathrm{e}^{\frac{t}{2}}+\mathrm{e}^{-\frac{t}{2}}}{2}\right)^s\right]\\
&=a^s \mathrm{e}^{r s}\left[\left(\frac{\mathrm{e}^x-\mathrm{e}^{-x}}{2 x}\right)^s-\lambda+(\lambda-1)\left(\frac{\mathrm{e}^x+\mathrm{e}^{-x}}{2}\right)^s\right] (\text {作变换 } t=2 x)\\&=a^s \mathrm{e}^{x s}\left[\left(\frac{\operatorname{sh} x}{x}\right)^s-\lambda+(\lambda-1)(\operatorname{ch} x)^s\right]
\end{aligned}
\]
设 $f(x)=\left(\frac{\operatorname{sh} x}{x}\right)^s-\lambda+(\lambda-1)(\operatorname{ch} x)^s$, 其中 $s>0, x>0$.
由引理知， $1<\frac{\operatorname{sh} x}{x}<\operatorname{ch} x(x>0), x \operatorname{ch} x-\operatorname{sh}$ $x>0$，所以
\[
\begin{aligned}
f'(x)&= s\left(\frac{\operatorname{sh} x}{x}\right)^{s-1} \cdot \frac{x \operatorname{ch} x-\operatorname{sh} x}{x^2}+ s(\lambda-1)(\operatorname{ch} x)^{s-1} \cdot \operatorname{sh} x \\
&< s(\operatorname{ch} x)^{s-1} \cdot \frac{x \operatorname{ch} x-\operatorname{sh} x}{x^2}+ s(\lambda-1)(\operatorname{ch} x)^{s-1} \cdot \operatorname{sh} x \\
&= \frac{s(\operatorname{ch} x)^{s-1}}{x^2}\left[x \operatorname{ch} x-\operatorname{sh} x+(\lambda-1) x^2 \operatorname{sh} x\right] .
\end{aligned}
\]
设
\[
g(x)=x \operatorname{ch} x-\operatorname{sh} x+(\lambda-1) x^2 \operatorname{sh} x, x>0,
\]
则
\[
\begin{aligned}
g'(x)&= \operatorname{ch} x+x \operatorname{sh} x-\operatorname{ch} x+(\lambda-1)(2 x \operatorname{sh} x+x^2 \operatorname{ch} x) \\
&= x[(2 \lambda-1) \operatorname{sh} x+(\lambda-1) x \operatorname{ch} x] .
\end{aligned}
\]
对于（1），当 $\lambda \leqslant \frac{2}{3}$ 时，由引理知 $x \operatorname{ch} x>\operatorname{sh} x$，
又 $\lambda-1<0$，所以 $(\lambda-1) x \operatorname{ch} x<(\lambda-1) \operatorname{sh} x$，
又 $\operatorname{sh} x>\operatorname{sh} 0=0$，因此
\[
\begin{aligned}
g'(x) & <x[(2 \lambda-1) \operatorname{sh} x+(\lambda-1) \operatorname{sh} x] \\
& =(3 \lambda-2) x \operatorname{sh} x \leqslant 0.
\end{aligned}
\]
从而 $g(x)$ 在 $(0,+\infty)$ 上单调递减，所以 $g(x)<g(0)=0$，故 $f^{\prime}(x)<0$，从而 $f(x)$ 在 $(0,+\infty)$ 上单调递减．
显然，$\operatorname{ch} 0=1$．由洛必达法则，知
\[
\begin{aligned}
f(0) & =-\lambda+(\lambda-1) \times 1+\lim _{x \rightarrow 0}\left(\frac{\operatorname{sh} x}{x}\right)^s \\
& =-1+\left(\lim_{x \to 0}\operatorname{ch} x\right)^s=-1+1=0
\end{aligned}
\]
所以 $f(x)<f(0)=0$，故 $L^s<\lambda G^s+(1-\lambda) A^s$ ． |
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kuing
Posted at 2023-10-27 14:47:10
isee
Posted at 2023-10-30 21:45:04
![233536wiq6z462rbkbz5la[1].png](data/attachment/forum/202503/23/210011wboqm1s90qx508b3.png "233536wiq6z462rbkbz5la[1].png")