Does this inequality about the second derivative of logarithm of modified Bessel function hold? In this paper, I came across the statement:

It is easy to confirm that$$\Bigg(\frac{I'_{\frac{D}{2}-1}(t)}{I_{\frac{D}{2}-1}(t)}\Bigg)'\le 0$$ for all $t > 0$ and $D \ge 2$.

For $I_\nu$ the modified Bessel of first kind, order $\nu$.
By the quotient rule and a few Bessel identities:
$$\begin{align}
\Bigg(\frac{I'_{\nu}(t)}{I_{\nu}(t)}\Bigg)' &= \frac{I''_\nu(t)I_\nu(t) - (I'_\nu(t))^2}{(I_\nu(t))^2} \\
&=\frac{I''_\nu(t)}{I_\nu(t)} - \Big(\frac{I'_\nu(t)}{I_\nu(t)}\Big)^2 \tag{1}\\
&= \frac{1}{4}\frac{I_{\nu-2}(t) + 2I_{\nu}(t) + I_{\nu+2}(t)}{I_\nu(t)} - \Big(\frac{I_{\nu+1}(t) + \frac{\nu}{x}}{I_\nu(t)}\Big)^2
\end{align}$$
For $\nu=0$: 
$$\begin{align}
\Bigg(\frac{I'_{0}(t)}{I_{0}(t)}\Bigg)' &= \frac{1}{4}\frac{ 2I_{0}(t) + 2I_{2}(t)}{I_0(t)} - \Big(\frac{I_{1}(t)}{I_0(t)}\Big)^2 \\
&= \frac{1}{2} + \frac{I_2(t)}{2I_0(t)} - \Big(\frac{I_{1}(t)}{I_0(t)}\Big)^2
\end{align}$$
Plots of this function show it greater than zero; I've tried several variations of $(1)$ using recurrence formulas, all equivalent and suggesting the inequality holds for orders greater than zero.
Where have I erred?
 A: You can continue with your approach, although using $2I'_{\nu} = I_{\nu+1}+I_{\nu-1}$ is easier: you can then group the terms into three expressions of the form
$$ I_{\mu+1}(x)I_{\mu-1}(x)-I_{\mu}^2(x), $$
which can be shown to be negative by expanding and using the Cauchy product of the series for each Bessel function (See this page of the paper V.R. Thiruvenkatachar, T.S. Nanjundiah, Inequalities concerning Bessel functions and orthogonal polynomials, Proc. Ind. Acad. Sci. Sect. A 33 (1951) 373–384).
However, I believe one can proceed with a similar approach directly. We have
$$ I_{\nu}(x) = \sum_{k=0}^{\infty} \frac{c_{\nu,k}}{k!} \left( \frac{x}{2} \right)^{k+2\nu}, $$
where $c_{\nu,k} = 1/\Gamma(\nu+k+1)$. Then
$$I_{\nu}'(x) = \sum_{k=0}^{\infty} \frac{(2k+\nu)c_{\nu,k}}{k!} \left( \frac{x}{2} \right)^{2k+\nu-1} \\
I_{\nu}''(x) = \sum_{k=0}^{\infty} \frac{(2k+\nu)(2k+\nu-1)c_{\nu,k}}{k!} \left( \frac{x}{2} \right)^{2k+\nu-2}, $$
so we have
$$ I_{\nu}''(x)I_{\nu}(x)-I_{\nu}'(x)^2 = \sum_{n=0}^{\infty}  \left( \frac{x}{2} \right)^{2n+2\nu-2} \sum_{k+l=n} \frac{2k+\nu}{k!n!}\big( (2l+\nu-1)-(2l+\nu) \big) c_{\nu,k}^2. $$
But every term on the right is nonpositive since the bracket reduces to $-1$.
