若a,b均为正实数,x,y∈R,且a+b=1,求证:ax^2+by^2>=(ax+by)^2
4个回答

x,y∈R

(x-y)2≥0

ab(x2+y2-2xy)≥0

1-a=b,1-b=a

abx2+bay2-2abxy≥0

a(1-a)x2+b(1-b)y2-2abxy≥0

(ax2-a2x2)+(by2-b2y2)-2abxy≥0

ax2+by2-(a2x2+2abxy+b2y2)≥0

ax2+by2-(ax+by)2≥0

ax2+by2≥(ax+by)2

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