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).


1 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 $a=2,b=1$

  • $\begingroup$ Wow. Thank you. It was so simple. I began well and got distracted by another problem which had a constraint. $\endgroup$ Jul 25, 2015 at 19:13
  • $\begingroup$ yes i agree you can come in my Mathcircle in Leipzig, you are wellcome $\endgroup$ Jul 25, 2015 at 19:17
  • $\begingroup$ @Dr.SonnhardGraubner Can I come ? $\endgroup$ Jan 28, 2017 at 10:06
  • $\begingroup$ do you have a question? right now $\endgroup$ Jan 28, 2017 at 10:07
  • $\begingroup$ @Dr.SonnhardGraubner No, I'd like to pay a visit to the Mathcircle in Leipzig $\endgroup$ Jan 28, 2017 at 10:08

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