I am a scientist who is trying to come up with some analytic solutions for a system that I only have approximate answers to and I have run into the problem of proving that the following function is monotone.
$$\sqrt{\lambda} \left[ \frac{I_0(2\sqrt{\lambda})}{I_1(2\sqrt{\lambda})} - \frac{I_1(2\sqrt{\lambda})}{I_0(2\sqrt{\lambda})} \right].$$
Where $I_n(x)$ is modified Bessel function of the first kind.
I am honestly not sure where to begin and any help or direction would be much appreciated!
On
As Robert Israel's answer says, we can reduce this to proving the inequality $$ x( I_1(x)^4 - I_0(x)^4) + 2 I_0(x)^3 I_1(x) < 0 ; $$ a more useful form for our purposes is $$ 1 - \left( \frac{I_1(x)}{I_0(x)} \right)^4 - \frac{2}{x} \frac{I_1(x)}{I_0(x)} > 0 . $$ It suffices to find a simple upper bound for the ratio, then verify it satisfies the inequality. The easiest way we are aware of follows Nåsell (1979), Rational bounds for ratios of modified Bessel functions, SIAM J. Math. Anal., Vol. 9, No. 1. (N.B. in this paper $(x)_k$ is the rising factorial and $[x]$ is the largest integer not larger than $x$, i.e. the floor of $x$.)
The idea is to use the recurrence relations $$ I_{\nu-1}(x) - I_{\nu+1}(x) = \frac{2\nu}{x} I_{\nu}(x) \\ I_{\nu-1}(x) + I_{\nu+1}(x) = 2I_{\nu}'(x) $$ repeatedly, to write derivatives of the expression $ g_{\nu}(x) = x^{-\nu} e^{-x} I_{\nu}(x) $ in terms of $I_{\nu}$ and $I_{\nu+1}$, and then use that $ (-1)^k g_{\nu}^{(k)}(x) $ is positive for $x>0$ and $\nu>-1$) (Nåsell does this by using that $g_{\nu}$ is the Laplace transform of a positive function); this expression can then be rearranged to find rational bounds for the ratio.
For example, we have $$ \begin{align} -g_{\nu}'(x) &= \nu x^{-\nu-1} e^{-x} I_{\nu}(x) + x^{-\nu} e^{-x} I_{\nu}(x) - x^{-\nu} e^{-x} I_{\nu}'(x) \\ &= x^{-\nu} e^{-x} \left( \left(\frac{\nu}{x}+1\right) I_{\nu} + \frac{\nu}{x} I_{\nu}(x) - I_{\nu-1}(x) \right) \\ &= x^{-\nu} e^{-x} \left( \left(\frac{2\nu}{x}+1\right) I_{\nu} - I_{\nu-1}(x) \right) ; \end{align}$$ so $$ \frac{2\nu+x}{x} I_{\nu} - I_{\nu-1}(x) > 0 , $$ or changing the index, $$ \frac{I_{\nu+1}(x)}{I_{\nu}(x)} > \frac{x}{2(\nu+1)+x} . $$
EDIT: the previous bound was not good enough since I made a calculation error. The idea still works, but is sadly rather more complicated that I originally hoped.
The bound Nåsell calls $U_{0,5,0}$ turns out to be sufficent, but easier to see is $$ \frac{I_1(x)}{I_0(x)} < U_{0,6,0} = \frac{ 2x(30 + 36 x + 27 x^2 + 16 x^3 + 16 x^4) }{ 120 + 144 x + 123 x^2 + 82 x^3 + 48 x^4 + 32 x^5 }, $$ which gives $$ \begin{align} &1 - \left( \frac{I_1(x)}{I_0(x)} \right)^4 - \frac{2}{x} \frac{I_1(x)}{I_0(x)} \\ &> \frac{x^2}{(120 + 144 x + 123 x^2 + 82 x^3 + 48 x^4 + 32 x^5)^4} \\ &\quad \times (25920000 + 124416000 x + 263044800 x^2 + 412784640 x^3 + 538137432 x^4 + 620215488 x^5 \\ &\qquad + 627520365 x^6 + 542714256 x^7 + 421454952 x^8 + 293542656 x^9 + 178465536 x^{10} \\ &\qquad + 92432384 x^{11} + 40183808 x^{12} + 16711680 x^{13} + 3670016 x^{14} + 524288 x^{15}), \end{align} $$ which is clearly positive since every coefficient is.
This is not an answer, but may be helpful as a start.
First of all, we might as well define $x = 2 \sqrt{\lambda}$. Then with $g(x) = I_0(x)/I_1(x)$, you want $x (g(x) - 1/g(x))$ to be monotone (decreasing) for $x > 0$. This turns out to be equivalent to $$ - {{ I}_{0}\left(x\right)} ^{4}x+ {{ I}_{1 }\left(x\right)} ^{4}x+2\, {{ I}_{0}\left(x\right)} ^{3}{{ I}_{1}\left(x\right)} < 0 $$
It's certainly true when $x$ is small and when $x$ is large (by using the first couple of terms of the known Maclaurin series and asymptotics of $I_0$ and $I_1$). With some effort it should be possible to translate those into explicit bounds: say $0 < x < \epsilon$ and $x > N$. That leaves a finite interval: by careful numerical evaluation at a finite number of points, plus bounds on derivatives, it should be possible to cover this interval by subintervals on which the inequality is true.