As part a much larger proof in my dissertation research (I am EE not mathematics) I am trying to show the following and have been stuck for weeks:
Suppose that $a > 0$ is real and let $f : (a,\infty) \to \mathbb{R}$ be defined by $$ f(x) = x \ln{\frac{x-a}{x+a}} $$ I am trying to show that $f(x) < -2a$ for all $x \in (a,\infty)$ but I keep getting stuck and getting caught going around in circles.
Plotting $f$ clearly shows that it is negative, is strictly increasing, and that $$ \lim_{x \to \infty} f(x) = -2a \,. $$ Being unable to show this inequality in any direct way, my strategy was to show the above limit, which is easy, and the fact that it is strictly increasing. The result readily follows and I am able to prove the more general theorem that shows this. The problem that has me going around in circles is showing that $f$ is strictly increasing. I tried showing this directly with no success, and trying to show that $f' > 0$ only results in having to show a more complicated inequality, which is the circular part. I could very well be missing some other obvious approach.
Any help with this would be greatly appreciated and it goes without saying that anyone who is able to help show this will receive credit in the final dissertation.
Following Kelenner's comment:
You have to show that
$$x \cdot \ln{\dfrac{x-a}{x+a}} \lt -2a$$
which is equivalent to
$$\ln{\dfrac{x-a}{x+a}}+\dfrac{2a}{x}<0$$
Let $g(x)=\ln{\dfrac{x-a}{x+a}}+\dfrac{2a}{x}$
Then the first derivative is $g'(x)=\dfrac{2a^3}{x^2(x-a)(x+a)}>0$, for all $x>a$
So $g$ s increasing in $(a,+\infty)$. But we also have:
$$\lim\limits_{x\to a^+}g(x)=-\infty$$ and
$$\lim\limits_{x\to+\infty}g(x)=0 $$
and also $g$ is obviously continuous in $(a,+\infty)$.
So $$g((a,+\infty))=(-\infty,0)$$
which means $g(x)<0$ , for all $x>a$ $\ \ \checkmark$