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Let $a$ and $b$ be positive real numbers. Show that
$$\frac{1}{a b} \geq \frac{4}{(a+b)^{2}}$$
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Cross multiplying both sides, it is equivalent to show that
$$(a+b) \cdot(a+b)=(a+b)^{2} \geq 4 a b \text { . }$$
To prove this, we apply AM-GM on both of the terms on the LHS:
$$(a+b) \cdot(a+b) \geq 2 \sqrt{a b} \cdot 2 \sqrt{a b}=4 a b \text { . }$$
by Diamond (75,914 points)

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