Question

已知a,b,x,y属于R,且a+b=1，求证（ax+by)(ay+bx)>=xy

Answer 1

a+b=1
(a+b)²=1
a²+2ab+b²=1
a²+b²=1-2ab

(ax+by)(ay+bx)
=a²xy+abx²+aby²+b²xy
=ab(x²+y²)+(a²+b²)xy
=ab(x²+y²)+(1-2ab)xy
=ab(x²+y²)-2abxy+xy
=ab(x²-2xy+y²)+xy
=ab(x-y)²+xy

∵a、b∈R+，即ab＞0，且(x-y)²≥0
∴ab(x-y)²≥0，即ab(x-y)²+xy≥xy

∴(ax+by)(ay+bx)≥xy

Answer 2

(ax+by)(ay+bx)
=(ax+(1-a)y)(ay+(1-a)x)
=(ax-x-ay)(ay-y-ax)
=((a-1)x-ay)((a-1)y-ax)>xy