Question

If x, y, z and w are four positive real numbers such that xyzw = 1 , what is the minimum value(1+x)(1+y)(1+z)(1+w)

Answer 1

From AM GM property,
$(x + y + z + w) \div 4 \geqslant \sqrt[4]{xyzw}$
or,
x+y+z+w >= 4
or
(1+x)+(1+y)+(1+z)+(1+w) >=4+1+1+1+1=8
again from AM GM property,
the maximum value will be, 2^4 = 16