Unable to use Solve to solve simultaneous equations, or use Maximize for exact maximum solution

Mathematica has found that the maximum of the following function is negative using NMaximize, but I am trying to prove that the maximum is negative (i.e. that the function is negative for all parameter values) - since it is an important step in a proof that I am writing:

Maximize[{(-1 + E^(δ ρ)) λ + (-1 +  E^(δ λ) + E^(δ ρ) - E^(δ (2 λ + ρ))) ρ, ρ > 0 && δ > 0 && λ > 0}, {ρ, λ, δ}]

However, Mathematica will not run this maximization, and it quickly returns the input as its output.

As an alternative method, I have tried to get Mathematica to simultaneously solve the first-order conditions of the maximization problem:

Solve[{-1 + E^(δ λ) - E^(δ (2 λ + ρ)) (1 + δ ρ) +  E^(δ ρ) (1 + δ (λ + ρ)) == 0, ρ (E^(δ λ) λ + E^(δ ρ) (λ + ρ) - E^(δ (2 λ + ρ)) (2 λ + ρ)) == 0, -1 + E^(δ ρ) + E^(δ λ) δ ρ - 2 E^(δ (2 λ + ρ)) δ ρ == 0}, {ρ, δ, λ}]

The Solve operation has been running on a HPC for a few hours now - does anyone have any advice as to how to make this run faster? I have been given that the advice that "approximate" solutions found through NMaximize would not suffice for the proof.

Thank you

-- I have also tried putting this through Matlab, but I am getting a lot of error messages.

Not a complete answer but an observation that may help to to solve your actual problem.

The expression

a = (-1 + E^(δ ρ)) λ + (-1 + E^(δ λ) + E^(δ ρ) - E^(δ (2 λ + ρ))) ρ

is nonpositive for all nonnegative δ, ρ, λ if and only if the following expression is nonnegative for all nonnegative x, y:

b[x_,y_] = δ a /. {ρ -> x/δ, λ -> y/δ} // Simplify

-(-1 + E^y) (-1 + E^x + E^(x + y)) x + (-1 + E^x) y

That reduces the number of variables by one. Might be helpful.

Edit:

The following should prove your claim:

The above substitution reduces the problem to showing that $$b(x,y) \leq 0$$ for all $$x \geq 0$$ and all $$y\geq 0$$, where

$$b(x,y) := \left(e^x-1\right) y-x \left(e^y-1\right) \left(e^{x+y}+e^x-1\right).$$

We have

$$b(x,0) = 0$$

and

$$\frac{\partial b}{\partial y}(x,0) = e^x (1-2 x)+x-1 \leq 0$$

for all $$x \geq 0$$. Moreover,

$$\frac{\partial^2 b}{\partial y^2}(x,y) = x \left(-e^y\right) \left(4 e^{x+y}-1\right) \leq 0$$ for all $$x \geq 0$$ and $$y \geq 0$$.

Hence we obtain $$\frac{\partial b}{\partial y}(x,y) = \frac{\partial b}{\partial y}(x,0) + \int_0^y \frac{\partial^2 b}{\partial y^2}(x,t) \, \mathrm{d} t \leq \frac{\partial b}{\partial y}(x,0) \leq 0$$

for all $$x \geq 0$$, $$y\geq 0$$ and

$$b(x,y) = b(x,0) + \int_0^y \frac{\partial b}{\partial y}(x,t) \, \mathrm{d} t \leq b(x,0) = 0.$$

• Thank you - this looks rather promising, I think it will take me a bit to process the argument. I'm not quite sure that I understand the first step though - can you please explain how showing that the second expression is nonnegative shows that the first expression is nonnegative? – Elke Oct 17 '18 at 13:54
• That's quickly done: We multiply the first expression by $\delta \geq 0$, so this does not change signs. Moreover we only rescale $\lambda$ and $\varrho$ by $\delta \geq 0$, so their signs will also not be flipped. If you like, you can treat the case $\delta = 0$ first. Then expression a equals 0. – Henrik Schumacher Oct 17 '18 at 13:58