Showing that $\coth(A-t)+\coth(t-B)$ is a concave function for $AGiven constants $A$ and $B$, with $A<B$, show that the function $\coth(A-t)+\coth(t-B)$ is concave function for $A<t<B$. Is it true?
 A: First we recognize that a function $f(x)$ is concave over an interval if and only if $\frac{d^2}{dx^2}f(x)<0$ over that interval. Let $g(t)=\coth(A-t)+\coth(t-B)$; therefore, we can compute the second derivative of $g$ to show whether it is concave.
Finding the first derivative gives: $$\frac{d}{dt}g(t)=\frac{d}{dt}[\coth(A-t)+\coth(t-B)]\\=\frac{d}{dt}\coth(A-t)+\frac{d}{dt}\coth(t-B)\\=\operatorname{csch}^2(A-t)-\operatorname{csch}^2(t-B)$$
Taking the derivative again gives: $$\frac{d^2}{dx^2}g(x)=\frac{d}{dx}[\operatorname{csch}^2(A-t)-\operatorname{csch}^2(t-B)]\\=\frac{d}{dx}\operatorname{csch}^2(A-t)-\frac{d}{dx}\operatorname{csch}^2(t-B)\\=2\operatorname{csch}^2(A-t)\coth(A-t)+2\operatorname{csch}^2(t-B)\coth(t-B)$$
Remember, to show concavity of $g$, we need to show that $\frac{d^2}{dt^2}g(t)<0$ over the interval $t\in(A,B)$, so let us continue with a proof by contradiction: assume that the opposite is true, $$2\operatorname{csch}^2(A-t)\coth(A-t)+2\operatorname{csch}^2(t-B)\coth(t-B)\geq0,t\in(A,B)$$ Because $\operatorname{csch}^2(A-t)$ is always positive, we can divide both sides of the inequality without worrying about the sign.$$\coth(A-t)+\frac{\operatorname{csch}^2(t-B)}{\operatorname{csch}^2(A-t)}\coth(t-B)\geq0$$ For simplicity, let $c(t)=\frac{\operatorname{csch}^2(t-B)}{\operatorname{csch}^2(A-t)}$, which is always positive; $$\therefore c(t)\coth(t-B)\geq-\coth(A-t)\\c(t)\coth(t-B)\geq\coth(t-A)$$ However, on the interval $t\in(A,B)$, it follows that $t-B<0$ and $t-A>0$, $$\therefore\coth(t-B)<0\\\therefore c(t)\coth(t-B)<0$$ The second statement follows from $c(t)>0$, but importantly, it also follows that $$\coth(t-A)>0$$ However, we just concluded that $$c(t)\coth(t-B)\geq\coth(t-A)$$ So something strictly less than zero is greater than or equal to something strictly greater than zero? This is clearly absurd. This must mean that our assumption was wrong. Therefore, is must be that $$2\operatorname{csch}^2(A-t)\coth(A-t)+2\operatorname{csch}^2(t-B)\coth(t-B)<0,t\in(A,B)\\\therefore\frac{d^2}{dt^2}g(t)<0,t\in(A,B)$$ which, by definition, means $g$ is concave over $t\in(A,B)$, and therefore the original equation of $\coth(A-t)+\coth(t-B)$ is concave over the interval $t\in(A,B)$, $\Bbb{QED}$.
