If a,b,c,d are 4 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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a

b

c

d

a b c d

=

1

(1 + a) (1 + b) (1 + c) (1 + d)

a b c d = 1

(1 + a) (1 + b) (1 + c) (1 + d) \geq 16

If a, b, c are positive real numbers such that a + b + c = 1 then the least value of $\frac{(1 + a) (1 + b) (1 + c)}{(1 - a) (1 - b) (1 - c)} i s$ is

a, b, c, d

a b c d = 1

(a + 4) (b + 4) (c + 4) (d + 4)

a, b, c, d

a + b + c + d = \frac{4}{3}

\sqrt{\frac{1}{b + c + d}} + \sqrt{\frac{1}{c + d + a}} + \sqrt{\frac{1}{d + a + b}} + \sqrt{\frac{1}{a + b + c}}

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