<s>[INST] An estimate for dyadic numbers I would like to prove that for some positive $\delta<1/2$ we have the following inequalities
$$ |\frac{1}{2^{n+1}} - \frac{1}{2^{m+1}}| \leqslant \delta\left( |\frac{1}{2^{n}} - \frac{1}{2^{m+1}}| + |\frac{1}{2^{m}} - \frac{1}{2^{n+1}}|\right)\qquad (m,n\in \mathbb{N})$$
$$ |\frac{1}{2^{n+1}}+i\frac{1}{2^{n+1}}  - \frac{1}{2^{m+1}}| \leqslant \delta\left( |\frac{1}{2^{n}} + i\frac{1}{2^{n}}- \frac{1}{2^{m+1}}| + |\frac{1}{2^{m}} - \frac{1}{2^{n+1}}- i\frac{1}{2^{n+1}}|\right).$$
I tried to make $n$ a continuous variable and use calculus but failed with computations. Is there any way to see those inequalities?
I would be happy with an existential result for $\delta$.
 [/INST] you can take $\delta=\frac{1}{3}$.
for any $a,b$ we have
$$\left|a-b\right|=\frac{1}{3}\left|3a-3b\right|=\frac{1}{3}\left|\left(2a-b\right)+\left(a-2b\right)\right|$$
now using triangle inequality and noting that $\left|a-2b\right|=\left|2b-a\right|$ we get
$$\le\frac{1}{3}\left(\left|2a-b\right|+\left|2b-a\right|\right)$$
now set $b=\frac{1}{2^{m+1}}$ whereas a in the first is $\frac{1}{2^{n+1}}$ and in the second $a=\frac{1+i}{2^{n+1}}$. 
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