Need to find least value of an algebraic expression without helper constraints.

Question

I am trying to solve this problem:

Given $a>b>0$, find the least value of $a + \frac {1}{b(a-b)}$

Initially I was confused and things got better when I re-wrote $a + \frac {1}{b(a-b)}$ as $(a-b) + b + \frac {1}{b(a-b)}$

Then, the given equation assumed a friendlier form $x + y + \frac{1}{xy} $

While further searching this community, I found a very similar problem here. This problem was finding the minimum value of $a + b + \frac{1}{ab}$. But there was a constraint helping this case: $a^2+b^2=1$. They established $(2+\sqrt2)$ as the minimum value.

I followed similar lines (AM>GM inequality) and ended up with a complex expression that looked like $2b(a-b) \le a^2+2b^2-2ab$ as there was no helper constraints.

I wonder (for the second time today) whether the problem statement is incomplete or a minimum value can be reached without constraints such as $a^2+b^2=1$.

Could you please help me out here?

Note: I know the Calculus way of solving this problem. But I am supposed to use algebra (inequalities).

Answer 1

i think you gave us the hint by yourself:$$a+\frac{1}{b(a-b)}=a-b+b+\frac{1}{b(a-b)}\geq 3\sqrt[3]{\frac{(a-b)b}{b(a-b)}}=3$$ by AM-GM generated with no Mathematica. The equal sign holds for $a=2,b=1$