Proving Monotonicity Elegantly

Question

What are some slick ways to prove, with respect to $x\in[0,\infty)$, $S_1(x)$ is increasing, and given $b\in[0,\infty)$, $S_2(x)$ is decreasing as defined below? I have proved it by taking several rounds of derivatives for various functions in the intermediate steps. That is not very elegant. \begin{align} S_1(x) &:=\frac{x}{2}\frac{1+e^{-x}}{1-e^{-x}}, \\ S_2(x) &:=\Big(\frac{x}{1-e^{-x}}\Big)^2\Big(e^{-x}+be^{-2x}\Big)=\Big(\frac{x}{e^x-1}\Big)^2(e^x+b), \\ x&\in[0,\infty), \quad b\in[0,\infty). \end{align} Would some clever integration over some parameters other than $x$ of some functions of certain signs do the trick?

Answer 1

@mickep's answer is good. I just came up with a proof that may be called elegant and is just the kind using integration with a parameter that I am seeking.

For $S_1$: $$\frac1{S_1(2x)} = \frac1x\tanh(x) = \int_0^1 \mathrm{sech}^2(tx)\,\mathrm dt.$$ $\cosh(tx)$ increases over $x$ for every $t\ge0$, so the integral decreases over $x$. Thus $S_1(x)$ increases.

For $S_2$: \begin{align} S_2 &= \frac4{f\big(\frac{x}2\big)^2}+\frac b{g(x)^2}, \\ f(x) &= \frac{e^x-e^{-x}}{x}=2\frac{\sinh(x)}x=2\int_0^1\cosh(tx)\,\mathrm dt, \\ g(x) &= \frac{e^x-1}x=\int_0^1e^{tx}\,\mathrm dt. \end{align} We see both $f$ and $g$ are positive and increasing in two ways. One way is that we observe both integrands above are positive and increase over $x$ for every $t\ge0$. The second way is that when Tayler expanding $f$ and $g$, we observe they have only non-negative power terms and all the coefficients are positive. Thus $S_2$ decreases.

Answer 2

I don't know if you consider this elegant, but anyways, here is a way not differentiating too much:

We start with $S_1$

For later use, we note that $$ S_1(x)=\frac{x}{2}\cdot\coth(x/2). $$ Differentiating $S_1$, we find $$ S_1'(x)=\frac{\sinh x-x}{2(\cosh x-1)}. $$ Since $D\sinh x=\cosh x$ and $D\cosh x=\sinh x$ and moreover $\sinh 0=0$ and $\cosh 0=1$, we find that $S_1'(x)>0$ for $x>0$ since, if we expand both the numerator and the denominator into Maclaurin series the first term will cancel the $x$ (in the numerator) and the $1$ (in the denominator) and all other terms will be positive.

Next, we consider $S_2$

We will write $S_2(x)=\psi(x)+b\phi(x)$, where $\psi$ and $\phi$ will be introduced below. We let $$ \psi(x):=\Bigl(\frac{x}{e^x-1}\Bigr)^2e^x=\Bigl(\frac{x}{2}\text{csch}(x/2)\Bigr)^2 $$ and hence its derivative is $$ \psi'(x)=-\frac{x}{2}\Bigl(\frac{x}{2}\coth(x/2)-1\Bigr)\text{csch}^2(x/2). $$ Note that since we have $S_1(x)-1$ in the parenthesis of $\psi'(x)$, and we just checked that $S_1'(x)>0$ for $x>0$, we conclude that (since $S_1(x)\to 1$ as $x\to 0$) that $\psi'(x)<0$ for $x>0$.

Next, let $\phi(x)=e^{-x}\psi(x)$. Then, since $\psi$ is a positive and decreasing function, it follows that $$ \phi'(x)=e^{-x}\bigl(\psi'(x)-\psi(x)\bigr)<0. $$ Since $$ S_2(x)=\psi(x)+b\phi(x) $$ and both $\psi$ and $\phi$ decrease, and $b\geq 0$, we conclude that $S_2$ is decreasing.