Showing $(4H-2L)e^{\frac{H+L}{k}}-2(H-L)(e^{H/k}+e^{\frac{H+2L}{k}})-L(e^{2H/k}+e^{2L/k})\geq 0$ for $k>0$, $H>0>L$, and $H+L>0$

Question

Let $k>0$, $H>0>L$, and $H+L>0$. Prove that $$(4H-2L)e^{\frac{H+L}{k}}-2(H-L)(e^{H/k}+e^{\frac{H+2L}{k}})-L(e^{2H/k}+e^{2L/k})\geq 0.$$

I'm trying to prove the statement in the title algebraically, but seems to be missing some tricks necessary to do so. I know that as $k$ goes to infinity, the requirement that $H+L>0$ is binding. I'm positive the statement is true through Mathematica visualizations, but am struggling with the formal proof. Any suggestions or tricks are welcome!

Answer 1

It suffices to prove that $$(4H-2L) - 2(H-L)(\mathrm{e}^{-L/k} + \mathrm{e}^{L/k}) - L(\mathrm{e}^{(H-L)/k} + \mathrm{e}^{-(H-L)/k}) \ge 0.$$ Denote the LHS by $f(k)$. We have \begin{align*} f'(k) &= \frac{(L-H)L}{k^2} (2\mathrm{e}^{-L/k} - 2\mathrm{e}^{L/k} - \mathrm{e}^{(H-L)/k} + \mathrm{e}^{-(H-L)/k}) \\ &\le \frac{(L-H)L}{k^2} (2\mathrm{e}^{-L/k} - 2\mathrm{e}^{L/k} - \mathrm{e}^{(-L-L)/k} + \mathrm{e}^{-(-L-L)/k})\tag{1}\\ &= \frac{(L-H)L}{k^2} (2\mathrm{e}^{-L/k} - 2\mathrm{e}^{L/k} - \mathrm{e}^{-2L/k} + \mathrm{e}^{2L/k})\\ &= \frac{(L-H)L}{k^2}(\mathrm{e}^{-L/k} - \mathrm{e}^{L/k})(2 - \mathrm{e}^{-L/k} - \mathrm{e}^{L/k})\\ &\le 0 \end{align*} where in (1) we use $H > -L > 0$.

Also, $\lim_{k\to \infty} f(k) = 0$. Thus, $f(k) \ge 0$ for all $k > 0$.

We are done.