Find the minimum value of $1000a+\frac{1}{1000b(a-b)}$ Given that $a>b>0$ , find the least value of:- $$1000a+\frac{1}{1000b(a-b)}$$
Can anyone please help me out with this one? I tried but couldn't think of anything. Tried using A.M.G.M inequality but that gives an expression containing both $a$ and $b$ in the expression so it can't be used to find the least value i think. Please! Help! Thanks a lot!

Hint: For any fixed $a > 0$, the quantity $1000a+\dfrac{1}{1000b(a-b)}$ is minimized by making $b(a-b)$ as large as possible. What value of $b$ (in terms of $a$) does this?
After figuring that part out, you'll be left with a one variable problem, which can be done with AM-GM or basic calculus.
