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Home » Since $ g''(x) = \frac4x^3 > 0 $ for $ x > 0 $, this is a minimum. Thus, the minimum value of $ f(n) $ is $ \sqrt2 + \frac2{\sqrt2} = \sqrt2 + \sqrt2 = 2\sqrt2 $. - AMAZONAWS

Since $ g''(x) = \frac4x^3 > 0 $ for $ x > 0 $, this is a minimum. Thus, the minimum value of $ f(n) $ is $ \sqrt2 + \frac2{\sqrt2} = \sqrt2 + \sqrt2 = 2\sqrt2 $. - AMAZONAWS

📅 March 11, 2026 👤 scraface
Mar 11, 2026
Since $ g''(x) = \frac{4}{x^3} > 0 $ for $ x > 0 $, this is a minimum. Thus, the minimum value of $ f(n) $ is $ \sqrt{2} + \frac{2}{\sqrt{2}} = \sqrt{2} + \sqrt{2} = 2\sqrt{2} $.

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