已知a,b＞0,a+b=1,求证(ax+by)(ay+bx)≥xy

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+...