Question

已知a,b＞0,a+b=1,求证(ax+by)(ay+bx)≥xy

Answer 1

证明:

(ax+by)(ay+bx)-xy

=xya^2+abx^2+aby^2+xyb^2-xy

=xy(a^2+2ab+b^2)-2abxy+abx^2+aby^2-xy

=xy(a+b)^2+ab(x^2-2xy+y^2)-xy 因为a+b=1

=xy+ab(x-y)^2-xy

=ab(x-y)^2 又因为:abxy都是正实数

则ab(x-y)^2>=0

即(ax+by)(ay+bx)-xy>=0

也就是(ax+by)(ay+bx)>=xy