Bounding a modified Bessel function of the first kind. Let $I_0$ be the zeroth-order modified Bessel function of the first kind. We know that, asymptotically as $x\to \infty$, $I_0(x) \sim e^x/\sqrt{2\pi x}$. Does anybody have a reference for the maximum of $e^{-x} \sqrt{x} I_0(x)$ for $0\leq x\leq \infty$? (A plot shows that it's $0.4688...$, but of course that is no proof.)
 A: Here's a crude but rigorous bound, while we wait for the question to be settled definitively.
Let $x\geq 0$. By definition (and half-angle formulas),
$$I_0(x) = \frac{1}{\pi} \int_0^\pi e^{x \cos \theta} d\theta 
= \frac{2 e^x}{\pi} \int_0^{\pi/2} e^{-2 x \sin^2 \alpha} d \alpha,$$ 
where we do the change of variables $\alpha = \theta/2$. Now, for $\alpha\in
\lbrack 0, \pi/2\rbrack$, $\sin \alpha \geq 2 \alpha/\pi$. Hence,
$$\int_0^{\pi/2} e^{-2 x \sin^2 \alpha} d \alpha \leq
\int_0^{\pi/2} e^{- \frac{8 x}{\pi^2} \alpha^2} d\alpha <
\frac{\pi}{\sqrt{8 x}} \int_0^\infty e^{-t^2} dt = \frac{\pi^{3/2}}{2^{5/2} \sqrt{x}}.$$
Hence,
$$I_0(x) < \sqrt{\frac{\pi}{8}} \frac{e^x}{\sqrt{x}}$$ 
for all $x\geq 0$. Of course, $\sqrt{\pi/8} = 0.626657\dotsc$, so this is indeed crude in comparison with $0.468822\dotsc$.
What we really need is a bound better than $0.46882 e^x/\sqrt{x}$ for $x\geq 2$ (say), so that the problem gets reduced to that of maximizing the Bessel function on a compact interval (such as \lbrack 0,2\rbrack) where it can be estimated well by a truncated power series.
A: By just considering the derivative of $\sqrt{x}\,I_0(x)\,e^{-x}$ and setting it equal to zero we have that the maximum is attained in the only point for which:
$$f(x)=(1-2x)\,I_0(x)+2x\, I_1(x) = 0.\tag{1}$$
Such function is convex and decreasing over $(0,x_0=2.3555\ldots)$, concave and decreasing over $(x_0,+\infty)$. Since $f(0)=1$ and $f(x_0)<0$, we can just find the only zero $\xi$ of $f$ by applying the Newton's method with starting point $x=0$. After two iterations we get:
$$\xi\geq \frac{1}{2}+\frac{I_1(1/2)}{I_0(1/2)},$$
after five iterations we get:
$$\xi\approx 0.78997842\ldots $$
from which it follows that:
$$\sqrt{x}\,I_0(x)\,e^{-x}\leq 0.4688223555.$$
From now on, this will be the Helfgott's constant.
Addendum.
Since $I_n(z)=\frac{z}{2n}(I_{n-1}(z)-I_{n+1}(z))$ by the Briggs' formulas, we have the continued fraction representation:
$$\frac{I_1}{I_0}(z)=\frac{1}{\frac{2}{z}+\frac{1}{\frac{4}{z}+\frac{1}{\frac{6}{z}+\ldots}}}\tag{2}$$
hence, due to $(1)$, our stationary point is just the solution of the continued fraction equation:

$$\frac{1}{\frac{2}{z}+\frac{1}{\frac{4}{z}+\frac{1}{\frac{6}{z}+\ldots}}}=1-\frac{1}{2z}.\tag{3}$$

A: Here's a full answer, involving some (rigorous) computation towards the end.
Let $f(t) = e^{-t} I_0(t) \sqrt{t}$, as above. Let us start by
showing that $f(t) < 0.45168$ for $t>8$. (This part of the lemma's proof was graciously provided by G. Kuperberg.)
First of all,
$$f(t) = \frac{e^{-t}}{\pi} \sqrt{t} \int_0^\pi e^{t \cos \theta} d\theta
= \frac{2}{\pi} \sqrt{t} \int_0^1 \frac{e^{-2 t s^2}}{\sqrt{1-s^2}} ds,$$
where we substitute $\cos \theta = 1 - 2 s^2$, and note that
$d\theta = 2 ds/\sqrt{1-s^2}$. We break the integral on $s$ in two at
$s=1/2$. 
The integrand from $0$ to $1/2$ is majorized by $(1.03+s^2/2) e^{-2 t s^2}$ 
(because
$1/\sqrt{1-s^2} - s^2/2$ has positive derivative for $s\geq 0$, and
$1/\sqrt{1-(1/2)^2} - (1/2)^2/2 = 1.0297\dotsc$); the integrand from $1/2$
to
$1$ is majorized by $e^{-t/2}/\sqrt{1-s^2}$. The first integral can be
extended from $0$ to $\infty$; the second one is exactly $e^{-t/2}\cdot \pi/3$.
We conclude that 
$$\begin{aligned}
f(t) &\leq \frac{2 \sqrt{t}}{\pi} \left(1.03\cdot \frac{\sqrt{\pi/2}}{
2 \sqrt{t}}
+ \frac{1}{2} \frac{\sqrt{\pi/2}}{8 t^{3/2}}\right) + 
\frac{2 \sqrt{t} e^{-t/2}}{3}\\
&\leq \frac{1.03}{\sqrt{2 \pi}} + \frac{1}{8 \sqrt{2\pi} t} + 
\frac{2 \sqrt{t} e^{-t/2}}{3} \leq 0.45168\end{aligned}$$
for $t\geq 8$.
In the range $0\leq t\leq 8$, we will use the Taylor expansion around
$t=0$. By (9.6.12) in Abramowitz-Stegun,
$$\begin{aligned}
I_0(t) = \sum_{n=0}^\infty \frac{(t^2/4)^2}{n!^2} \leq
\sum_{n=0}^{15} \frac{(t^2/4)^n}{n!^2} + 
\frac{16^{16}}{16!^2} \cdot \sum_{m=0}^\infty \frac{1}{17^m} 
\leq \sum_{n=0}^{15} \frac{(t^2/4)^n}{n!^2} + 0.00000005.\end{aligned}$$
We now use the bisection method (with $25$ iterations), implemented via interval arithmetic
(as described in, e.g., \S 5.2 of Tucker, Validated Numerics), to ascertain that
the maximum of the series $\sum_{n=0}^{15} (t^2/4)^n/n!^2$ on the interval
$\lbrack 0,8\rbrack$ lies between $0.46882234$ and $0.46882237$. End of proof!
