Is it known, and if yes how does one show, that the function
$$ \psi(n):=n\int_0^{+\infty} e^{-x}I_0\left(\frac{x}{n}\right)^{n-1}I_1\left(\frac{x}{n}\right)\mathrm{d}x$$
is decreasing for all $n\ge 3$? Here $I_0$ and $I_1$ denote the modified Bessel functions of the first kind of order $0$ and $1$ respectively.
As $I_0^\prime = I_1$, we have $ I_0\left(\frac{x}{n}\right)^{n-1}I_1\left(\frac{x}{n}\right) = \frac{d}{dx} \left(I_0(\frac{x}{n})^n\right)$ , so $$ \psi(n) = - n + n\int_0^\infty I_0\left(\frac{x}{n}\right)^n\exp(-x)dx,\\ = -n +n^2 \int_0^\infty (I_0(y) \exp(-y))^n dy $$
and as $0\leq \exp(-x)I_0(x)\leq1$, expanding $(n+1)^2$, $$ \psi(n+1)-\psi(n) = - 1 + (n+1)^2 \int_0^\infty (I_0(y) \exp(-y))^{n+1} dy \\ -(n)^2 \int_0^\infty (I_0(y) \exp(-y))^{n} dy \\ \leq (2n+1)\int_0^\infty (I_0(y) \exp(-y))^{n+1} dy-1. $$ Now let us use a rough upper bound for $I_0(x) \exp(-x)$, e.g., $$ I_0(x) \exp(-x) \leq \frac{4}{3\sqrt{2\pi x+1}}. $$ Then $$ \frac{4}{3}(2n+1)\int_0^\infty(2\pi x+1)^{-(n+1)/2}=\frac{4}{3\pi}\frac{2n+1}{n-1} $$ and provided $n\geq 10$, $$ \psi(n+1)-\psi(n) \leq 0 $$
