一个指数函数的分式的单调性

hbghlyj

在Show an inequality by Vasile Cirtoaje中提到

Let $a>b>d$, $a>c>d$, $s<t$ and $s,t\neq0$ then we have:
$$\frac{e^{as}-e^{bs}}{e^{cs}-e^{ds}}<\frac{e^{at}-e^{bt}}{e^{ct}-e^{dt}}$$
The inequality can be found in the Dictionary of Inequalities second edition by Professor Peter Bullen (2015) p85.
The inequality above is a consequence of the MVT.

等价于$\displaystyle\frac{e^{as}-e^{bs}}{e^{at}-e^{bt}}<\frac{e^{cs}-e^{ds}}{e^{ct}-e^{dt}}$.
由Cauchy中值定理,存在$ξ∈(b,a),η∈(d,c)$满足\begin{gathered}\frac{e^{as}-e^{bs}}{e^{at}-e^{bt}}=\frac ste^{(s-t)ξ}\\\frac{e^{cs}-e^{ds}}{e^{ct}-e^{dt}}=\frac ste^{(s-t)η}\end{gathered}但是没法证明$ξ>η$

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Cauchy's mean value theorem
If the functions $f$ and $g$ are both continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists some $c\in (a,b)$, such that\[(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c).\]
Of course, if $g ( a ) ≠ g ( b )$ and $g'( c ) ≠ 0$, this is equivalent to:
$${\frac {f'(c)}{g'(c)}}={\frac {f(b)-f(a)}{g(b)-g(a)}}.$$

hbghlyj

设$a>b>d≥0$, $a>c>d≥0$, $s<t$ 且 $s,t\neq0$ 我们要证明
$$\frac{e^{as}-e^{bs}}{e^{cs}-e^{ds}}<\frac{e^{at}-e^{bt}}{e^{ct}-e^{dt}}$$
等价于$$\frac{e^{(a-d)s}-e^{(b-d)s}}{e^{(c-d)s}-1}<\frac{e^{(a-d)t}-e^{(b-d)t}}{e^{(c-d)t}-1}$$
将$a-d,b-d,c-d$换元为$a,b,c$, 化为以下命题:
设$a>b>0$, $a>c>0$, $s<t$ 且 $s,t\neq0$ 则
$$\frac{e^{as}-e^{bs}}{e^{cs}-1}<\frac{e^{at}-e^{bt}}{e^{ct}-1}$$
设$x=e^{cs}$,化为$$\frac{x^{a/c}-x^{b/c}}{x-1}\text{关于$x$单增}$$
将$a/c,b/c$换元为$a,b$, 化为以下命题:
设$a>b>0$, $a>1$, $x>0$, 则
$$g(x)=\frac{x^a-x^b}{x-1}关于x单增\tag1$$

hbghlyj

设$f(t)=(tx-t-x) x^t\;,t>0$
则$f'(t)=x^t (1-x) \log (x) \left(\frac{x}{x-1}-\frac{1}{\log (x)}-t\right)$
因为$(1-x)\log(x)<0$与$0<\frac{x}{x-1}-\frac1{\log (x)}<1\;∀x>0$
所以$f(t)$在$t\in\left[0,\frac{x}{x-1}-\frac1{\log (x)}\right]$单减, 在$t\in\left[0,\frac{x}{x-1}-\frac1{\log (x)}\right]$单增.
因为$a>1$, 所以$a>\frac{x}{x-1}-\frac1{\log (x)}$.

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Case 1. 当$b>1$时, 有$a>b>1$,
所以$f(a)>f(b)$
所以$x(x-1)^2g'(x)=(a x-a-x) x^a-(b x-b-x)x^b=f(a)-f(b)>0$
从而$g'(x)>0$
(1)证毕.

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Case 2. 当$b<1$时, 有$f(b)<f(0)=-x$
和Case 1同样地, 只需要证明$f(a)>f(b)$, 那么只需要证明$f(a)>-x$, 即$(a x-a-x) x^{a-1}>-1$, 左边记为$h(x)$, 则$h'(x)=(a-1) a (x-1) x^{a-2}$, 所以$h(x)$在$[0,1]$单减,在$[1,+∞)$单增, 所以$h(x)\ge h(1)=-1$. (1)证毕.