Given:

    x,y,z > 0
    xy+yz+zx <= 3/4

and

    P=x+y+z+(1/x)+(1/y)+(1/z)

Find the Minimum Value of P. Find x, y, and z when P = Minimum Value.

Use Cauchy's inequality:
xy+yz+zx >= 3*(x^2*y^2*z^2)^(1/3)
=====>(x^2*y^2*z^2)^(1/3) x*y*z = 15/((4^12)*(x*y*z)^3))^(1/15)
==> P >= 15/((4^12)*(1/8^3))^(1/15)
==> P >= 15/((2^24)/(2^9))^(1/15)
==>P >= 15/(2^15)^(1/15)=15/2
"=" x=y=z=1/2

Posted by vohonam on 2002-06-20 16:03:57