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Question 12

$$a, b$$ are positive real numbers such that $$\frac{1}{a} + \frac{9}{b} = 1$$. The smallest value of $$a + b$$ is

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By the Cauchy-Schwarz inequality, $$(a+b)\left(\frac{1}{a}+\frac{9}{b}\right) \geq (1+3)^2 = 16$$. Since $$\frac{1}{a}+\frac{9}{b}=1$$, this gives $$a+b \geq 16$$, with equality achievable, so the smallest value is 16.

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