For $d > 0$ and $\log x = \log_ex$ (second one is shortened version of log)
$$\frac{2d}{\exp(-2\log(2d))} \le \frac{1}{2d}$$
So far my steps are, from logarithmic property: $$\exp(-2\log(2d)) = \exp(\log(2d)^{-2}) = (2d)^{-2} = \frac{1}{4d^2} $$
And therefore: $$\frac{2d}{\exp(-2\log(2d))} = \frac{2d}{\frac{1}{4d^2}} = 8d^3$$ Which can't be less or equal than $\frac{1}{2d}$. What am I supposing wrong?
Since $e(\cdot)>0$ and $d>0$, you can rewrite
$$ e^{-2 \log 2d}\geq 4d^2 \implies \frac{1}{4d^2} \geq 4d^2 \implies (4d^2-1)(4d^2+1) \leq 0 $$ In $\mathbb{R}$ this has solution when $(2d-1)(2d+1) \leq 0$, which has a solution when $-\frac{1}{2} \leq d \leq \frac{1}{2}$, and now combine it together with the condition on $d$ that you have