An irresistible inequality The following occurred while working on some research project. Since the methods of proof I used were lengthy, I wish to see a skillful or insightful approach (perhaps even conceptual). Anyhow, here it is. Let 
$$f(x)=\left(\frac{x}{e^x-1}\right)^2 + \left(\frac{x+1}{e^{x+1}+1}\right)^2.$$
Can one give a short and elegant proof of these statements?
(1) $f(x)$ is a strictly decreasing function of $x$ over $\mathbb{R}$.
(2) In fact, the statement holds true if $e$ is replaced by any real number $t>1$.
 A: I do not know wether this helps you or not, but you may do as follows.
Denote $f(x)=(e^x-1)/x$. Note that $f'(x)=\frac{fe^x(x-1+e^{-x})}{x(e^x-1)}>0$, so
$f$ is a positive increasing function. 
Lemma. The function $1/f=x/(e^x-1)$ is a (positive decreasing) convex function.
Proof. 
$$
(1/f)''=\frac{e^x(2+x-(2-x)e^x)}{(e^x-1)^3},
$$
we have to check that it is non-negative. If $x\geqslant 2$, this is clear.
If $0<x<2$, this reduces to 
$$
e^x\leqslant \frac{2+x}{2-x}=1+x+x^2/2+x^3/2^2+x^4/2^3+\dots,\,\,(1)
$$
that holds coefficient-wise: $n!\geqslant 2^{n-1}$ for $n\geqslant 1$.
Finally, the case $x<0$ reduces to $x>0$, since $2+x-(2-x)e^x$ and
$2-x-(2+x)e^{-x}$ always have opposite signs.
Corollary. $g:=1/f^2=x^2/(e^x-1)^2$ is convex.
Proof. $-(1/f^2)'=2(-1/f)'(1/f)$, both multiples are positive decreasing functions,
thus their product also decreases.
Now we claim that $g(x-a)+x^2/(e^x+1)^2$ decreases for each $a\geqslant 0$,
for $a=1$ we get your statement (and for other $a$ something equivalent to your remark).
Since $g'$ increases, we have $g'(x-a)\leqslant g'(x)$, so it suffices to check this for $a=0$.
We have $g(x)+x^2/(e^x+1)^2=2x^2(e^{2x}+1)/(e^{2x}-1)^2$. Denote $2x=y$,
we need to check that $y^2(e^y+1)/(e^y-1)^2$ decreases. Taking logarithmic derivative,
this is equivalent to $$\frac{2}x+\frac{e^x}{e^x+1}-\frac{2e^x}{e^x-1}\leqslant 0.$$
For $x=-y>0$ we have $$\frac2{y}-\frac2{e^y-1}\geqslant \frac2y-\frac2{y+y^2/2}=\frac1{1+y/2}>\frac1{1+e^y}$$
as desired. For $x>0$ we have 
$$
\frac{2}x+\frac{e^x}{e^x+1}-\frac{2e^x}{e^x-1}\leqslant \frac{2+x}x-\frac{2e^x}{e^x-1}=
\frac{(2-x)e^x-(2+x)}{x(e^x-1)}\leqslant 0
$$
by (1).
