Inequality with decay of modified Bessel functions of the second kind I think that the following inequality holds for all $x > 0$ and all $\nu$ above some constant that is somewhere around 0.2:
$$
K_\nu(2 x) \le \frac{2^{2 - 2 \nu}}{\Gamma(\nu)} x^\nu K_\nu^2(x)
$$
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 $\nu$:

How can this inequality be proved? Ideally it'd include the lower bound on $\nu$ for which it becomes true for all $x$, but doing it just for $\nu \ge \frac12$ or $\nu > 1$ would also be helpful.
I've tried manipulating the integral representation $K_\nu(x) = \int_0^\infty \exp(- x \cosh t) \cosh(\nu t) \,\mathrm{d}t$, but didn't get anywhere.

$$K_\nu(z)=\frac{2^\nu\,\Gamma\left(\nu+\frac{1}{2}\right)}{z^\nu\sqrt{\pi}}\int_{0}^{+\infty}\frac{\cos(zt)}{(t^2+1)^{\nu+\frac{1}{2}}}\,dt $$
so we have to compare
$$ I_2=\int_{0}^{+\infty}\int_{0}^{+\infty}\frac{\cos(zt)\cos(zs)}{\left((t-s)^2+(1+ts)^2\right)^{\nu+\frac{1}{2}}}\,ds\,dt$$
and 
$$ I_1 = \int_{0}^{+\infty}\frac{\cos(2zu)}{(u^2+1)^{\nu+\frac{1}{2}}}\,du.$$
It is reasonable to expect that the main contribute to $I_2$ is given by the points close to the line $s=t$.
That should give an approximate duplication formula for $K_\nu$ as wanted. Continues.
