If a,b,c & d are four positive real numbers such that abcd=1 , what is the minimum value of (1+a)(1+b)(1+c)(1+d).

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$We have, a b c d = 1 Now, (1 + a) (1 + b) (1 + c) (1 + d) = 1 + a + b + c + d + a b + a c + a d + b c + b d + c d + a b c + b c d + a c d + a b d + a b c d = 1 + (a + b + c + d) + (a b + a c + a d + b c + b d + c d) + (a b c + b c d + a c d + a b d) + a b c d = 1 + (a + b + c + d) + (a b + a c + a d + \frac{1}{a d} + \frac{1}{a c} + \frac{1}{a b}) + (\frac{1}{d} + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + 1 (Using a b c d = 1) = 2 + (a + \frac{1}{a}) + (b + \frac{1}{b}) + (c + \frac{1}{c}) + (d + \frac{1}{d}) + (a b + \frac{1}{a b}) + (1 + \frac{1}{a c}) + (1 + \frac{1}{a d}) \geq 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 (x + \frac{1}{x} \geq 2, for x \geq 0) = 16 Hence, (1 + a) (1 + b) (1 + c) (1 + d) \geq 16$

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