For which $\alpha>0$ does $x\le|W(-cx^2)|^{-\alpha}$ Let $x\in(0,1)$. I want to know for which $\alpha>0$ it's true that
$$
x\le|W(-cx^2)|^{-\alpha},\label{1}\tag{$\ast$}
$$
where $W$ is the Lambert W-function and $c>0$ is some constant.
In my numerical tests, the value of $c$ didn't really seem to matter, but \eqref{1} seemed to hold for very small $\alpha$, for example $\alpha\approx 0.001$. It seems difficult to prove analytically because of the non-elementary nature of the Lambert W function.
For negative $y<0$, it seems to be true that $W(-y)<0$. So we can rewrite \eqref{1} as
$$
x(-W(-cx^2))^{-\alpha}\le 1.
$$
We can define a function $f(x)=x(-W(-cx^2))^{-\alpha}$. Then $f(0)=0$, $f>0$ on $(0,1)$ and $f\in C^1$ since $W$ is differentiable on $(0,1)$ as it does not include the points $\{0,\frac{1}{e}\}$.
So the maximum of $f$ reached at $x_0$ should satisfy
$$f'(x_0)=0\label{2}\tag{$\ast\ast$}$$
where
$$
f'(x)=\left(-W(-cx^2)\right)^\alpha\left(1-\frac{2 c\alpha x^2 W'(-cx^2)}{W(-cx^2)} \right),
$$
So \eqref{2} is
\begin{align*}
&\left(-W(-cx_0^2) \right)^\alpha-2c\alpha x_0^2\left(-W(cx_0^2) \right)^{\alpha-1}W'(-cx_0^2)=0
\\
\iff& \alpha=-c'\frac{W(-cx_0^2)}{x_0^2 W'(-cx_0^2)}
\\
\iff& x^2_0\frac{\mathrm{d}}{\mathrm{d}x_0}\log\left( W(-cx_0^2)\right)=-c''\frac{1}{\alpha},
\end{align*}
But I don't see how to go from here, i.e., how to invert the function
$$
\psi(x)=x^2\frac{\mathrm{d}}{\mathrm{d}x}\log\left(W(-cx^2)\right)
$$
to recover $x_0$ as
$$
x_0=\psi^{-1}\left(-c''\frac{1}{\alpha}\right),
$$
and plug that back into \eqref{1}.
But in Mathematica, it gives
$$
c'''\psi^{-1}\left(-c''\frac{1}{\alpha}\right)=\pm c'''\left(\alpha W\left(\mp c'' i\frac{1}{\sqrt{\alpha}}\right)\right)^{-\frac{1}{2}},
$$
which isn't very helpful!
 A: There are a few ambiguities:

*

*By treating $W(-cx^{2})$ as real-valued, you're implicitly assuming $cx^{2} \leq e^{-1}$.

*There are two different values for $W(-cx^{2})$, the principal branch in $(-1, 0)$ and the lower branch in $(-\infty, -1)$, which are you using?

*On the principal branch, your inequality trivially holds for all $\alpha > 0$ since $x <1 < |W(-cx^2)|^{-\alpha}$.

Thus, there are two possible nontrivial questions:

*

*For which $\alpha > 0$ does $x \leq |W(-cx^{2})|^{\alpha}$ hold for all $x \in (0, \min\{1, 1/\sqrt{ce}\})$ on the primary branch of $W$.

*For which $\alpha > 0$ does $x \leq |W(-cx^{2})|^{-\alpha}$ hold for all $x \in (0, \min\{1, 1/\sqrt{ce}\})$ on the lower branch of $W$.

My conclusions are as follow:

*

*The inequality holds if $\alpha \leq 1/2$ and $c$ is not too small.

*The inequality holds for any given $\alpha$ if $c$ is not too small.


Question 1
$W(-cx^{2})$ is defined to be the solution on $(-1,0)$ to the equation:
$$-c x^{2} = we^{w}$$
But what matters is $|W(-cx^2)|^{\alpha}$, so let's rewrite the equation in terms of $\hat{w}$ where:
$$\hat{w} = (-w)^{\alpha} \in (0,1)$$
$$ \Rightarrow w = -\hat{w}^{1/\alpha}$$
Thus, $|W(-cx^2)|^{\alpha}$ is the solution on $(0,1)$ to the equation:
$$c x^{2} = \hat{w}^{1/\alpha}e^{-\hat{w}^{1/\alpha}}$$
The right hand side is increasing for $\hat{w} \in (0,1)$, since its derivative is:
$$(1/\alpha)\hat{w}^{1/\alpha-1}e^{-\hat{w}^{1/\alpha}}(1-\hat{w}^{1/\alpha})$$
Therefore, the solution is at least $x$ if and only if:
$$cx^{2} \geq x^{1/\alpha}e^{-x^{1/\alpha}}$$
$$\Leftrightarrow \log(c) + (2-1/\alpha)\log(x) \geq - x^{1/\alpha}$$
$$\Leftrightarrow \log(c) \geq (1/\alpha - 2)\log(x) - x^{1/\alpha} \equiv \gamma(x)$$

If $\alpha > 1/2$, then the right hand is arbitrarily large for small $x$, hence the inequality is violated.
If $\alpha = 1/2$, then $\log(c) \geq -x^{2}$ only holds for all $x \in (0,1)$ if $c \geq 1$.
If $\alpha < 1/2$, then let's maximize $\gamma(x)$:
$$\gamma'(x) = \frac{1/\alpha - 2}{x} - (1/\alpha)x^{1/\alpha - 1}$$
$$\gamma''(x) < 0$$
The first order condition is $\gamma'(\tilde{x})=0 \Rightarrow \tilde{x} = (1 - 2\alpha)^{\alpha}$, so:
$$\begin{align}
\gamma(\tilde{x}) &= (1 - 2\alpha)\log(1-2\alpha) - (1 - 2\alpha) \\
&=(1-2\alpha)(\log(1-2\alpha)-1)
\end{align}$$
Thus, the inequality holds if:
$$\boxed{\log(c) \geq (1-2\alpha)(\log(1-2\alpha)-1)}$$
This is sufficient but not quite necessary since $\tilde{x}$ may lie outside of $(0, 1/\sqrt{ce})$. Instead, let $\tilde{x}(c) = \min\left\{(1 - 2\alpha)^{\alpha}, \, 1/\sqrt{ce}\right\}$, then the inequality holds for any $\alpha$ and $c$ that satisfy:
$$\log(c) \geq \gamma(\tilde{x}(c))$$
(We can solve for the exact bound on $c$ as a function of $\alpha$ by setting it equal.)

Question 2
$W(-cx^{2})$ is defined to be the solution on $(-\infty,-1)$ to the equation:
$$-c x^{2} = we^{w}$$
But what matters is $|W(-cx^2)|^{-\alpha}$, so let's rewrite the equation in terms of $\hat{w}$ where:
$$\hat{w} = (-w)^{-\alpha} \in (0,1)$$
$$ \Rightarrow w = -\hat{w}^{-1/\alpha}$$
Thus, $|W(-cx^2)|^{-\alpha}$ is the solution on $(0,1)$ to the equation:
$$c x^{2} = \hat{w}^{-1/\alpha}e^{-\hat{w}^{-1/\alpha}}$$
The right hand side is increasing for $\hat{w} \in (0,1)$, since its derivative is:
$$(1/\alpha)\hat{w}^{-1/\alpha-1}e^{-\hat{w}^{-1/\alpha}}(\hat{w}^{-1/\alpha}-1)$$
Therefore, the solution is at least $x$ if and only if:
$$cx^{2} \geq x^{-1/\alpha}e^{-x^{-1/\alpha}}$$
$$\Leftrightarrow \log(c) + (2+1/\alpha)\log(x) \geq - x^{-1/\alpha}$$
$$\Leftrightarrow \log(c) \geq -(1/\alpha + 2)\log(x) - x^{-1/\alpha} \equiv \hat{\gamma}(x)$$

There are no obvious cases here, so let's just maximize $\hat{\gamma}(x)$. The first order condition is characterized by:
$$\hat{\gamma}'(x) = -\frac{1/\alpha + 2}{x} + (1/\alpha)x^{-1/\alpha - 1}$$
$$\hat{\gamma}'(\tilde{x}) = 0 \Rightarrow \tilde{x} = (1+2\alpha)^{-\alpha}$$
$$\hat{\gamma}''(x) = \frac{1/\alpha + 2}{x^{2}} - (1/\alpha)(1/\alpha + 1)x^{-1/\alpha - 2}$$
$$\hat{\gamma}''(\tilde{x})<0$$
Since the first-order condition is only satisfied at $\tilde{x}$, and $\hat{\gamma}$ is concave at that point, $\hat{\gamma}$ is increasing on $(0,\tilde{x})$ and decreasing on $(\tilde{x}, 1)$.
$$\hat{\gamma}(\tilde{x}) = (1+2\alpha)(\log(1+2\alpha)-1)$$
Thus, the inequality holds if:
$$\boxed{\log(c) \geq (1+2\alpha)(\log(1+2\alpha)-1)}$$
This is sufficient but not quite necessary since $\tilde{x}$ may lie outside of $(0, 1/\sqrt{ce})$. Instead, let $\tilde{x}(c) = \min\left\{(1 + 2\alpha)^{-\alpha}, \, 1/\sqrt{ce}\right\}$, then the inequality holds for any $\alpha$ and $c$ that satisfy:
$$\log(c) \geq \hat{\gamma}(\tilde{x}(c))$$
(We can solve for the exact bound on $c$ as a function of $\alpha$ by setting it equal.)
