Is it true $\left|\dfrac{1}{2^a}-\dfrac{1}{2^b}\right|Let $0 \leq a <b<1$ be two real numbers
Is it true that
$$
\left|
\dfrac
{1}
{2^a}
-
\dfrac
{1}
{2^b}
\right|
<b
$$
 A: Here is another version $f(x)=\frac{1}{2^x}$. Using MVT, $\exists c\in(a.b)$ s.t.
$$\left|f(a)-f(b)\right|=\left|f'(c)\right|\left|a-b\right|$$
Or
$$\left|\frac{1}{2^a}-\frac{1}{2^b}\right|=\left|-\frac{1}{2^c} \ln{2}\right|\left|a-b\right|<\left|\frac{1}{2^c} \right|\left|b-a\right|=\frac{1}{2^c} (b-a)\leq b-a<b$$
A: I think it's true.
We need to prove that
$$b2^{a+b}+2^a-2^b>0$$ and since $2^a\geq1$, it's enough to prove that
$$b2^b+1-2^b>0$$ or $f(b)>0$, where
$$f(b)=b-1+2^{-b}.$$
Indeed, $$f'(b)=1-2^{-b}\ln2=\frac{2^b-\ln2}{2^b}>\frac{1-\ln2}{2^b}>0,$$
which says $$f(b)>f(0)=0$$ and we are done!
A: For fixed $b$, the maximum of the left side is when $a=0$, so you want:
$$1-\frac{1}{2^b}<b$$
for $b\in(0,1)$. (The absolute values are unnecessary.)
This is equivalent to:
$$2^b\leq \frac{1}{1-b}
$$
for $b\in(0,1)$.
This can be seen by Taylor series:
$$2^b=e^{b\log 2}=1+\sum_{k=1}^{\infty}\frac{\log^k 2}{k!}b^k$$
and
$$\frac{1}{1-b}=1+\sum_{k=1}^{\infty}b^k$$
But $0<\log 2<1$ so $\frac{\log^k 2}{k!}<\frac{1}{k!}\leq 1$ and thus $$2^b<\frac{1}{1-b}$$
A: As others have observed, it's enough to show that $1-{1\over2^b}\lt b$ for $b\gt0$.  We'll show a stronger inequality.
Since $b\ge0$, we have 
$${1\over 2^{bx}}=e^{-(b\ln2)x}\lt1$$ 
for $0\lt x$, and thus
$$1\gt\int_0^1e^{-(b\ln 2)x}dx={1\over b\ln2}{1\over2^{bx}}\Big|_1^0={1\over b\ln2}\left(1-{1\over2^b}\right)$$
which, since $0\lt\ln2\lt1$, gives us the stronger inequality
$$1-{1\over2^b}\lt b\ln2\quad\text{for }b\gt0$$
