I need to show that the following expression is always increasing.
$$\frac{I_1(x)}{x I_0(x)} + \left(\frac{I_1(x)}{I_0(x)}\right)^2 \quad x>0,$$
where the $I_n$ represent the modified Bessel functions of the first kind. Wolfram Alpha tells me this is true- however I can't think of any simple argument to show this is the case using pen and paper. Any help is appreciated.
We have the integral representation: $$ I_n(\alpha) = \frac{1}{\pi}\int_{0}^{\pi} \cos(nx)\,e^{\alpha\cos x}dx. \tag{1}$$ By considering the Taylor series of $e^y$ and the Fourier series of $(\cos x)^m$, we have that: $$ I_n(\alpha) = \sum_{\nu=0}^{+\infty}\frac{(\alpha/2)^{n+2\nu}}{\nu! (n+\nu)!}.\tag{2} $$ Moreover, integrating by parts and using Briggs' formulas we have: $$ I_n = \frac{\alpha}{2n}\left(I_{n-1}-I_{n+1}\right). \tag{3}$$ From this identity, we have that the ratio between $I_n$ and $I_{n-1}$ is the continued fraction: $$ \frac{I_n}{I_{n-1}}(\alpha) = \frac{1}{\frac{2}{\alpha}\,n + \frac{1}{\frac{2}{\alpha}\,(n+1) + \ldots}}.\tag{4}$$ In particular $$ r(x)=\frac{I_1}{I_{0}}(x) = \frac{1}{\frac{2}{x} + \frac{1}{\frac{4}{x} + \frac{1}{\frac{6}{x}+\frac{1}{\ldots}}}}= \frac{x}{2 + \frac{x^2}{4 + \frac{x^2}{6+\frac{x^2}{8+\ldots}}}}.\tag{5}$$ We also have $$\frac{d}{d\alpha}\,I_n = \frac{1}{2}\left(I_{n-1}+I_{n+1}\right)\tag{6} $$ hence it is not difficult to exploit $(5)$ and $(6)$ for proving that the derivative of the wanted function is positive on $\mathbb{R}^+$. You may also have a look at the applications of Turàn's inequality to modified Bessel functions.
