Inequal or Equal?

By the AM-GM inequality,

$\large a + \dfrac{1}{b(a-b)}=(a-b)+b+\dfrac{1}{b(a-b)} \ge 3 \sqrt[3]{(a-b).b.\dfrac{1}{b(a-b)} } =3$

Equality holds if $a-b=b=\dfrac{1}{b(a-b)}$ , i.e. , a= 2 and b= 1.

Thus the answer is 3.

While it doesn't hold for all of problems of this type, it was possible to solve this one logically. If a is greater than b and both of them must be positives, then what are the least integers that a and b can take on? a = 2 and b = 1. Then it becomes clear how the answer is 3. I really don't think that we should rely on hard mathematical truths, instead we should think about what problems give to us and solve it using our minds. ;)

a=b+k with k>0.

The problem turns into:

S=b+k+1/bk>=3 (\sqrt_3 (b×k×1/bk))

So S>=3

When S=3: k=b=1/bk

So k=b=1, a=2.

Answer: 3