Two Bessel Functions 
Let $I_k$ denote the modified Bessel function of the first kind of index $k$. Show that, if $y > x > 0$, then
  $$-I_0(x)I_1(y) + I_0(y)I_1(x) <0$$

I am looking for a source that provides this information. 
 A: From the integral representation
$$
I_0(x)I_0(y)=\frac{1}{\pi}\int_0^\pi I_0(\sqrt{x^2+y^2-2xy\cos\theta})d\theta,
$$
using the formula $I_1(x)=I_0'(x)$ we immediately see that
$$-I_0(x)I_1(y)+I_0(y)I_1(x)=\frac{1}{\pi}\int_0^\pi \frac{I_1(\sqrt{x^2+y^2-2xy\cos\theta})}{\sqrt{x^2+y^2-2xy\cos\theta}}(x-y)(1+\cos\theta)d\theta.
$$
For $0<x<y$ this is clearly less than $0$, thus proving $$-I_0(x)I_1(y)+I_0(y)I_1(x)<0,~0<x<y.$$
Since $I_1(x)/x$ is monotonically increasing for $x>0$, it is also possible to prove a stronger inequality

$$
-I_0(x)I_1(y)+I_0(y)I_1(x)<-I_1(y-x),~0<x<y.
$$

A: There is a useful lemma that settles the question very nicely here. 

If $f(z)$ and $g(z)$ are two real analytic functions in a neighbourhood $U$
  of the origin, $$ f(z)=\sum_{n\geq 0}a_n z^n,\qquad g(z)=\sum_{n\geq
> 0}b_n z^n $$ with $a_n,b_n>0$ and the sequence
  $\left\{\frac{a_n}{b_n}\right\}_{n\geq 0}$ being an
  increasing/decreasing sequence, then the ratio
  $r(x)=\frac{f(x)}{g(x)}$ is an increasing/decreasing function over
  $U$.

The proof is not difficult: it is enough to consider $f'g-g\, f'$ in terms of power series to deduce that $\frac{dr}{dx}$ has constant sign. In our case, the coefficients of $I_0(x)$ and $I_1(x)$ have a simple structure, and the above lemma leads to:


*

*$\displaystyle\frac{I_0(x)}{I_1(x)}$ is a decreasing function over $\mathbb{R}^+$;

*$\displaystyle\frac{x\cdot I_0(x)}{I_1(x)}$ is an increasing function over $\mathbb{R}^+$.


Additionally, $\frac{I_0}{I_1}$ has a very nice representation in terms of continued fractions, that makes the above statements almost trivial.
