If $b\geq l$ with $b,l>1$ then $\exists c>0: l\geq cb$? Let $T>1$ be a real number and $a_1$ be the least common multiple of the positive integers $a_0$ and $a$ with $a,a_0>1$ and $T>a_1$. How to show that there exists a positive constant $c$ such that
$$\left \lfloor{T/a_1}\right \rfloor\geq c T/a \ ?$$
This result seems to be obvious. I know that 
$$T/a\geq\left \lfloor{T/a_1}\right \rfloor>T/a-1$$
and obviously if $\left\lfloor{T/a_1}\right \rfloor> T/a-1$ then there is a real number between them, which could be written as $c T/a $. I don't know how to show that though.
 A: You say:  "then there is a real number between them,($\lfloor \frac T{a_1}\rfloor$ and $\frac Ta - 1$) which could be written as cT/a. I don't know how to show that though." 
If we continue that thought:
$\lfloor \frac T{a_1}\rfloor > \frac {\lfloor \frac T{a_1}\rfloor +(\frac Ta-1)}{2}> \frac Ta-1$ so if we let $c =\frac {\lfloor \frac T{a_1}\rfloor +(\frac Ta-1)}{2}\div \frac Ta=\frac {a[\lfloor \frac T{a_1}\rfloor +(\frac Ta-1)]}{2T}$ that will do it.... but Martund has a simpler idea.
.....
If $T > a_1$ then $\frac T{a_1} > 1$ and $\lfloor \frac T{a_1}\rfloor \ge 1 = \frac aT \cdot \frac Ta$.
So ... give a peak at Martund's answer.
"$c = \frac aT$ does the job."
.......
Of course you can choose any $c; 0< c \le \frac aT$ and if you choose $c < \frac aT$ we have $c\frac Ta < 1 \le \lfloor \frac T{a_1}\rfloor$ an even stronger result.
Unless of course, we don't know that there are any $c; 0< c < \frac aT$.... which is the Archemedian principal.  Which is what we are trying to prove. I guess.
A: $$c=\frac{a}{T}$$
does the job.
