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Let $a>b$ be positive real numbers. What is the minimum value of
$$
a+\frac{1}{b(a-b)} ?
$$
in Mathematics by Diamond (75,914 points) | 19 views

1 Answer

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Best answer
We rewrite the expression as
$$
b+(a-b)+\frac{1}{b(a-b)}
$$
Applying AM-GM shows that
$$
b+(a-b)+\frac{1}{b(a-b)} \geq 3 \sqrt[3]{b \times(a-b) \times \frac{1}{b(a-b)}}=3
$$
Equality is achieved when $b=a-b=\frac{1}{b(a-b)}$, or when $a=2$ and $b=1$. Hence, the minimum of the expression is $3 . \square$
by Diamond (75,914 points)

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