prove that the following function is decreasing?

Question

I am trying to prove that the following function is decreasing. \begin{align}&f(t)=\frac{1-g(t)}{\sqrt{1+e^t}}\cdot\exp\left(-\frac{te^t}{2(1-e^t)}\right)&t<0\end{align}where $ g(t)=\dfrac{(1+e^t)(1+t-e^t)}{(1-e^t)^2}$. I have$$f^{\prime}(t)=-\frac{1}{2\sqrt{1+e^t}}\exp\left(-\frac{te^t}{2(1-e^t)}\right)\left[\frac{(1-g^2(t))e^t}{1+e^t}+\frac{2g(t)e^t}{1+e^t}+\frac{2(1+e^t)}{1-e^t}+\frac{4g(t)e^t}{1-e^t}\right],$$ but I can't show that $f^\prime(t)<0$. Can you help me?

Answer 1

If we run the calcs on what you wrote, we get \begin{align} &-2f'(t)\sqrt{1+e^t}\cdot\exp\left(\frac{te^t}{2-2e^t}\right)=\\ &=\frac{2e^{2t}}{(e^t-1)^4(e^t+1)}\left(2\cosh(2t)-t^2\cosh t-4t\sinh t+4\cosh t-t^2-6\right)\end{align}

After a small argument for the case $t=0$ (it turns out that $\lim_{t\to 0} f(t)$ and $\lim_{t\to 0} f'(t)$ exist) we realise that, to prove that $f$ is strictly decreasing on the whole real line, it is sufficient to prove that $$2\cosh(2t)-t^2\cosh t-4t\sinh t+4\cosh t-t^2-6\ge 0$$ and that the above expression has finitely many zeros.

If we call that function $h(t)$, we have

\begin{align} \boxed{h(t)}&=2\cosh(2t)-t^2\cosh t-4t\sinh t+4\cosh t-t^2-6\\&=\sum_{k=0}^\infty \frac{2^{2k+1}}{(2k)!}t^{2k}-\sum_{k=0}^\infty \frac{1}{(2k)!}t^{2k+2}-\sum_{k=0}^\infty \frac{4}{(2k+1)!}t^{2k+2}+\sum_{k=0}^\infty\frac{4}{(2k)!}t^{2k}-t^2-6\\&\boxed{=\sum_{n=0}^\infty a_nt^n} \end{align}

With $$\boxed{a_n=\begin{cases} 0&\text{if }n\le 2\text{ or }n\text{ is odd}\\ \dfrac{2^{n+1}}{n!}-\dfrac{1}{(n-2)!}-\dfrac{4}{(n-1)!}+\dfrac{4}{n!}&\text{ if }n\ge 4\text{ and }n\text{ is even} \end{cases}} $$

But $$\dfrac{2^{n+1}}{n!}-\dfrac{1}{(n-2)!}-\dfrac{4}{(n-1)!}+\dfrac{4}{n!}=\\=\frac{1}{n!}\left(2^{n+1}-n^2-3n+4\right)=\frac{1}{n!}(2^{n+1}-(n+4)(n-1))$$

And you can easily show that $2^{n+1}\ge (n+4)(n-1)$ for all $n\ge 4$. For instance by observing that the inequality holds for $n=4$ and that, for $n\ge 4$, $$\frac{(n+5)n}{(n+4)(n-1)}=\left(1+\frac{1}{n+4}\right)\left(1+\frac{1}{n-1}\right)\le \frac98\cdot\frac43=\frac32<2$$

Hence, all the $a_n$-s are $\ge 0$.

So, obviously $h(0)=0$ and $h(t)>0$ for all $t>0$.

But $h$ is an even function, hence $h(t)> 0$ for negative values of $t$ as well.