Find the smallest value of $$a + \frac {1}{(a-b)b} $$ where a>b>0
I found this question in AM-GM inequality problems but I am stuck at this
2026-04-08 12:31:37.1775651497
Find the Smallest Value
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Hints:
1) $a>b$ denote $$a=b+x, x \geqslant 0$$ 2) Prove $$f= b+x+\frac{1}{bx} \geqslant 3$$