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 9xyz≥4(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)+4−1xy+yz+zx
≥a((x+y+z)23(xy+yz+zx))+4−1xy+yz+zx (by the extended Cauchy-Schwarz inequality)
=a(13(xy+yz+zx))+4−1xy+yz+zx
=a−33(xy+yz+zx)+4
≥a−33((x+y+z)23)+4 since xy+yz+zx≤(x+y+z)23
=a−3+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