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?