If $a>b>0$, find the minimum value of $a+\cfrac{1}{(a-b)b}$

Question

I am clueless on how to simplify the expression $a+\cfrac{1}{(a-b)b}$.

My book gives as a hint to rewrite the given expression so that there's an $a-b$ in a numerator ,which can be then canceled with the denominator when AM-GM is applied,however I don't see how to do that.

André Nicolas · Accepted Answer · 2016-01-05T16:32:56.280

4

Let $a-b=c$. So we are looking at $b+c+\frac{1}{cb}$. By AM/GM this is $\ge 3$. The value $3$ is attained when $b=c=1$.

Remark: If we don't want to introduce $c$, we can rewrite our expression as $(a-b)+b+\frac{1}{(a-b)b}$. This feels less natural.

edited Jan 05 '16 at 16:32

answered Jan 05 '16 at 16:27

André Nicolas

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It's always a pleasure receiving an answer from you.Thanks for your time :) – Mr. Y Jan 05 '16 at 16:35
You are welcome. – André Nicolas Jan 05 '16 at 16:36

If $a>b>0$, find the minimum value of $a+\cfrac{1}{(a-b)b}$

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