A closed form solution for this exponential sum inequality $e^{-Ax} + e^{-Bx} \leq C$? Is there a closed form solution for $e^{-Ax} + e^{-Bx} \leq C$ where $A,B\in\mathbb{R}_{+}$ and $C\in [0,1]$ ?
 A: WLOG, assume $A > B > 0$ and $0 < C \le 1$. Let $p = \frac{B}{A} \in (0, 1)$.
Let $a = C^{p-1}$. Let $u = \frac{1}{C}\mathrm{e}^{-Ax}$.
We need to solve the equation $u + a u^p = 1$ which admits an infinite series solution (see [1])
$$u = \sum_{k=0}^\infty \frac{\Gamma(pk+1)a^k (-1)^k}{\Gamma((p-1)k+2) k!}.$$
Thus, the solution of $\mathrm{e}^{-Ax} + \mathrm{e}^{-Bx} = C$ is given by
$$x = - \frac{\ln C}{A} -\frac{1}{A}\ln \left(\sum_{k=0}^\infty \frac{\Gamma(pk+1)a^k(-1)^k}{\Gamma((p-1)k+2) k!}\right).
\tag{1}$$
For example, $A = \sqrt{5}, B = \sqrt{2}$, $C = \frac{2}{3}$,
(1) gives $x \approx 0.619497866$.
Reference
[1] Nikos Bagis, Solution of Polynomial Equations with Nested Radicals, https://arxiv.org/pdf/1406.1948.pdf
A: You have better to recast the equation into a symmetric form, by putting
$$
\left\{ \matrix{
  s = \left( {A + B} \right)/2 \hfill \cr 
  d = \left( {A - B} \right)/2 \hfill \cr}  \right.\quad  \Leftrightarrow \quad 
\left\{ \matrix{
  A = s + d \hfill \cr 
  B = s - d \hfill \cr}  \right.
$$
so as to get
$$
e^{\, - Ax}  + e^{\, - Bx}  = e^{\, - sx} \left( {e^{\, - dx}  + e^{\,dx} } \right)
 = 2e^{\, - sx} \cosh (dx)
$$
and so
$$
\cosh (dx) \le {C \over 2}e^{\,sx} 
$$
Then you can perform on this the various approximation processes already indicated.
A: I prefer to add another answer instead of adding to the previous one which is already too long.
Instead of the previous starting point, let us use
$$x_0=\frac{2\log \left(\frac{2}{c}\right)}{a+b} $$ which is obtained by the first iteration of Newton method starting at $x=0$. By Darboux theorem, this is an underestimate of the solution; its advantage is that it takes into account both $a$ and $b$.
The results for the previous four cases ($x_1$ being the first iterate of Newton method starting at $x_0$).
$$\left(
\begin{array}{cccccc}
a & b & x_0 & x_1 &   \text{solution} \\
 \pi    & e & 0.473148142 & 0.474869150 &  0.474869172 \\
 2 \pi  & e & 0.308015202 & 0.347822293 &  0.348346335 \\
 \pi  & \frac{e}{2} & 0.616030405 & 0.695644586 &    0.696692669 \\
 2 \pi  & 2 e & 0.236574071 & 0.237434575 &    0.237434586
\end{array}
\right)$$ The results are much better.
For the case where $b=\frac a2$, the ratio $\frac {x_1}{x_{sol}}$  starts at $1$ for $c=0$, goes through a minimum of $0.996777$ around $c=0.04$ and grows up to $0.999935$ for $c=1$. Much better again.
For the same case, using one iteration of Halley method, the ratio $\frac {x_1}{x_{sol}}$  starts at $1$ for $c=0$, goes through a maximum of $1.00091$ around $c=0.01$ and grows up to $1$ for $c=1$. Much better again.
For the same case, using one iteration of Householer method, the ratio $\frac {x_1}{x_{sol}}$  starts at $1$ for $c=0$, goes through a maximum of $1.000001$ around $c=0.21$ and grows up to $1$ for $c=1$. Much better again.
