An algebra problem by Jayanta Mandi

$a + \frac{1}{b(a-b)} =(a-b)+b+\frac{1}{b(a-b)}$

Now applying $AM \geq GM$ $a + \frac{1}{b(a-b)} =(a-b)+b+\frac{1}{b(a-b)} \geq 3\sqrt[3]{(a-b) b \frac{1}{b(a-b)}} =3$

The equality holds if $a-b=b=\frac{1}{b(a-b)}$

It is ideal that you rewrite the expression as @Jayanta Mandi has done. This is

$b+(a-b)+\dfrac{1}{b(a-b)}$

By the AM-GM Inequality,

$(b+(a-b)+\dfrac{1}{b(a-b)})/3 \geq \sqrt[3]{b(a-b)\dfrac{1}{b(a-b)}}$

Notice how in the RHS, the $b(a-b)$ cancels out. This leaves

$(b+(a-b)+\dfrac{1}{b(a-b)})/3 \geq \sqrt[3]{1}$

Multiplying by $3$ and rearranging we see that

$a+\dfrac{1}{b(a-b)} \geq \boxed{3}$

Great problem! :D

A calc bashing solution:

Differentiate our expression with respect to $a$ and get

$1-\frac{1}{b(a-b)^2}$

and setting to zero gives $b(a-b)^2=1$ (note this is the equality situation in Jayanta's solution). Solving, $a=b+\frac{1}{\sqrt{b}}$ and plugging in our expression we wish to minimize we want to minimize

$b+\frac{2}{\sqrt{b}}$

We can proceed by differentiating again or by using AM-GM. Either way, we find the expression is minimized when $b=1$ and our answer is 3.

to find the minimum numbers. assume a =2, b =1 ( since a>b>0)

2+(1/1) = 3

I don't do the sum by theorem. I think, If a>b>0 then b=1 (1 is greater than 0) and a=2 (2 is greater than 1). Then I sit the value of a,b and I found 2.Then I think how this problem is in level 3!! hehehe:):)