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+xz >= 3*(x^2*y^2*z^2)^(1/3)
==>(x^2*y^2*z^2)^(1/3) xyz 1/(xyz)^3 >= 8^3
1/x = 4/x +4/x +4/x +4/x
1/y and 1/z . Do the same as 1/x
Use Cauchy's inequality for 15 numbers
we have
P >= 15/((4^12)*(xyz)^3))^(1/15)
P >= 15*((1/xyz)^3 * (1/4^12))^(1/15)
==> P >= 15*(8^3 / 4^12)^(1/15)
==> P >= 15* (2^9 / 2^24)^(1/15)
==> P >= 15*(1/2^15)^(1/15)
P>=15/2
"=" x=y=z=1/2
If you write on paper, You can look better. Because this website, I can't post my solution by picture. I hope you understand.

Posted by vohonam on 2002-06-20 16:25:23