Decay relationship with modified Bessel functions of the second kind I think that the following inequality holds for all $x > 0$ and all $\nu \ge \frac{1}{2}$:
$$
K_\nu(2 x) \le \frac{2^{2 - 2 \nu}}{\Gamma(\nu)} x^\nu K_\nu^2(x)
,$$
where $K$ is a modified Bessel function of the second kind.
Here's a plot of $\frac{2^{2 - 2 \nu}}{\Gamma(\nu)} x^\nu K_\nu^2(x) - K_\nu(2 x)$ for different $\nu$s:

And the minimum of these curves over 10,000 points $0 < x \le 10$ while varying $\nu$ (which is of course just the curve's value at 10):

I've proved the relationship when $\nu$ is a half-integer on math.stackexchange; I'd like to show it for all other $\nu \ge \tfrac12$. So showing that the derivative in $\nu$ is always positive would be sufficient, but that derivative is gross: Letting $L_\nu(x) := \frac{\partial}{\partial \nu} K_\nu(x)$ and $\psi$ be the digamma function, the derivative is
$$
4 x^{\nu } K_{\nu }(x) \left(K_{\nu }(x) \left(\psi(\nu )+\log \left(\frac{4}{x}\right)\right)-2 L_\nu(x)\right)+4^{\nu } \Gamma (\nu ) L_\nu(2 x)
.$$
Neither I nor Mathematica can prove that it's positive – though it seems numerically like it is. Is there some duplication relation for $L_\nu$ or other property that will help prove this? Or a completely different approach?
 A: The proposer’s inequality can be improved by a factor of two, that is, 
$ (I) \quad K_\nu(2x) \le \, \dfrac{2^{1-2\nu}}{\Gamma(\nu)} \,x^\nu K_\nu^{\,2}(x) \,\,  ,x>0, \nu\ge 1/2 .$
Two integral identities and one inequality is needed:
$ (a) \quad K_\nu^{\,2}(x) =\frac{1}{2}\int_{0}^\infty\dfrac{dt}{t} e^{\large-t/2-x^2/t}
K_\nu(x^2/t)$
$ (b) \quad K_\nu(2x) =\frac{1}{2}x^\nu\int_{0}^\infty\dfrac{dt}{t^{\nu+1}} e^{\large-t-x^2/t}$
$ (c) \quad x^\nu K_\nu(x)>2^{\nu-1} \Gamma(\nu)\, e^{\large-x}$
The references are (a) DLMF 10.32.18 with x=y; (b) DLMF 10.32.10 with 2x as the argument; (c) J. Math. Anal. Appl. 420 (2014) 373-386.  It should be noted that the author of (c) claims that the region of validity is for x>0, $\nu > 0,$ but it appears that the proper region is for $\nu\ge 1/2 .$  (The derivation uses a formula that is good only for $\nu\ge 1/2,$ and plots indicate that the inequality breaks down for $\nu\lt 1/2$. Besides, the problem requires only $\nu\ge 1/2$.)  All formulas are valid for $x>0.$
In $(I)$, use $(a)$ on the RHS (right-hand side).  Use $(c)$ within the resulting integral to get 
$ \dfrac{2^{1-2\nu}}{\Gamma(\nu)} \,x^\nu K_\nu^{\,2}(x) \ge 
\dfrac{(2x)^\nu}{2}\int_{0}^\infty\dfrac{dt}{t^{\nu+1}} e^{\large-t/2-2x^2/t}
\Bigl(\dfrac{t}{2x}\Bigr)^{2\nu} \,.$
Now the RHS of the previous equation is, by (b), $K_{-\nu}(2x)=K_{\nu}(2x),$ where the well-known index reflection formula has been used. $\square$ 
