Find the minimum value of $1000a+\frac{1}{1000b(a-b)}$ [duplicate]
Solution 1:
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.
Solution 2:
Hint. Note that $$1000a+\frac{1}{1000b(a-b)}=1000(a-b)+1000b+\frac{1}{1000b(a-b)}$$ then apply AM-GM inequality (this time for three terms instead of two!).