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.

As x,y,z>0 min(x+1/x)=min(y+1/y)
Thus x=y=z (or x=1/y but xy+yz+zx<=3/4 so xy<=3/4)
so xy=x^2<=1/4
As 1/x increases exponentialy as x decreases linearly, min(x+1/x) occurs at max(x)=1/2
So P=3min(x+1/x)= 15/2

Posted by Nigel Nisbet on 2003-02-14 08:41:58