Processing math: 100%

Wednesday, April 27, 2016

Find, in terms of a, where a>0, the minimum value of the expression a(x2+y2+c2)+9xyzxy+yz+zx for all non-negative real x,y and z such that x+y+z=1.

Find, in terms of a, where a>0, the minimum value of the expression a(x2+y2+c2)+9xyzxy+yz+zx for all non-negative real x,y and z such that x+y+z=1.

Since 9xyz4(xy+yz+zx)1 (by the Schur's inequality), we can transform the objective function as

a(x2+y2+c2)+9xyzxy+yz+zx

=a(x2xy+yz+zx+y2xy+yz+zx+z2xy+yz+zx)+9xyzxy+yz+zx

a(x2xy+yz+zx+y2xy+yz+zx+z2xy+yz+zx)+4(xy+yz+zx)1xy+yz+zx (by the Schur's inequality)

=a(x2xy+yz+zx+y2xy+yz+zx+z2xy+yz+zx)+41xy+yz+zx

a((x+y+z)23(xy+yz+zx))+41xy+yz+zx (by the extended Cauchy-Schwarz inequality)

=a(13(xy+yz+zx))+41xy+yz+zx

=a33(xy+yz+zx)+4

a33((x+y+z)23)+4 since xy+yz+zx(x+y+z)23

=a3+4

=a+1

Therefore, the minimum of  a(x2+y2+c2)+9xyzxy+yz+zx is a+1, this occurs when x=y=z.

No comments:

Post a Comment