I want to prove that, given $a>1$, the function $$f(t)= \underbrace{e^{-\frac{2(a+t)}{1-t^2}}}_{f_1(t)} \underbrace{-\frac{2(a+t)}{1-t^2} + \frac 1{(1-t)^2} }_{f_2(t)}, \quad -1<t<0$$ has only one global maximum point. In fact, the function $f$ is near $-\infty$ for $t$ near $-1$, then it goes up to some point, then it goes down to $e^{-2a}-2a+1$ at $t=0$.
It is quite interesting to note that the two functions $f_1$ and $f_2$ have the same shape as that of the function $f$. To convince this point, I tried to draw the functions $f$, $f_1$, $f_2$ (in red, green, blue respectively) for $a=1$, $a=1.0125$, and $a=1.25$, and obtained the following pictures.
Apparently, the case $a=1$ is very different from the case $a>1$. The monotonicity of $f_1$ and $f_2$ can be easily proved because the derivatives of $f_1$ and $f_2$ are easy to examine. To be more precise, we have $$ f'_1(t)=-\frac{2(t^2+2at+1)}{(1-t^2)^2} f_1(t) $$ and $$ f'_2(t)=\frac{2t(t^2+2at+1 - 2(a-1))}{(1-t^2)^2(1-t)} . $$ However, the derivative of $f$ is quite hard since it is a linear combination of the exponential and fractional functions.
Another thing that I should mention is that if we denote by $t_i \in (-1,0)$ the critical point of $f_i$, with $i=1, 2$, namely $$t_1^2+2at_1+1=0=t_2^2+2at_2+1 - \underbrace{2(a-1)}_{>0},$$ then $$-1<t_1<t_2<0.$$
Please advise me if there is any trick to solve the question.