Find the minimum value of $1000a+\frac{1}{1000b(a-b)}$

Question

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!

score 5 · Accepted Answer · answered Jan 06 '22 at 09:20

5

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.

answered Jan 06 '22 at 09:20

JimmyK4542

54,331

1

This means that $b=a-b$ , which gives $b=0.5a$. – Ggsvdgdhddh Jan 06 '22 at 09:22

Robert Z · Answer 2 · 2022-01-06T09:54:22.293

5

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!).

edited Jan 06 '22 at 09:54

answered Jan 06 '22 at 09:26

Robert Z

145,942

Wow!!! Amazing one. So simple yet i couldn't see it. Thanks a lot!!!!!! – Ggsvdgdhddh Jan 06 '22 at 09:34

Find the minimum value of $1000a+\frac{1}{1000b(a-b)}$

2 Answers2