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Show that $\left(1+a^{2}\right)\left(1+b^{2}\right) \geq 4 a b$ for positive reals $a, b$.
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By AM-GM, we have $1+a^{2} \geq 2 a$ and $1+b^{2} \geq 2 b$. Hence,
$$\left(1+a^{2}\right)\left(1+b^{2}\right) \geq 2 a \times 2 b=4 a b$$
Note: We can expand the LHS, and then apply AM-GM directly on all the terms
by Diamond (75,914 points)

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