Inequality containing $x^2-2x^2e^{-ax}$

Question

Assuming there is an inequality in the form of

$x^2-2x^2e^{-ax}\leq \frac{b}{a^2}$

for $x,a,b>0$. Is it possible to find a bound for $x$?

I have difficulty obtaining the bound on $x$ because of the exponential term.

Answer 1

$f_a(x)=x^2(1-2e^{-ax})$ is quite obviously an unbounded function on $\mathbb{R}^+$, with a unique stationary point located at the positive root of $f_a'(x) = 2x((ax-2) e^{-ax}+1)$, which in terms of the Lambert W function is given by $$x_a=\frac{2-W(e^2)}{a}\approx \frac{31}{70a}.$$ $f_a(x_a)$ is an absolute minimum for $f_a(x)$ over $\mathbb{R}^+$ and $f_a(x)=0$ at $z_a=\frac{\log 2}{a}>x_a$. In particular $x\leq z_a$ clearly ensures $f_a(x)<0$, hence $f_a(x)\leq\frac{b}{a^2}$ for any positive $b$. In order to refine this inequality we may consider the Taylor series of $f_a(x)$ centered at $x=z_a$: $$ f_a(x) = \frac{\log^2 2}{a}(x-z_a)+\left(2\log 2-\frac{\log^2 2}{2}\right)(x-z_a)^2 +\left(a-a\log 2+\frac{a}{6}\log^2 2\right)(x-z_a)^3+\ldots$$ and get that $f_a(x)\leq\frac{b}{a^2}$ as soon as $x\leq z_a+K\frac{b}{a}$.