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求实数x,y的值使得(y-1)^2+(x+y-3)^2+(2x+y-6)^2取到最小值
人气:354 ℃ 时间:2019-08-16 22:20:09
解答
令a=y-1 b=x+y-3 c=2x+y-6
可得a-2b+c=-1
则原式=a^2+b^2+c^2
=a^2+1/4 b^2+1/4 b^2+ 1/4b^2 +1/4b^2+c^2
>=1/6(a-1/2b-1/2b-1/2b-1/2b+c)^2=1/6(a-2b+c)^2=1/6
取等条件a=-1/2b=c
即y-1 = -1/2 (x+y-3) = 2x+y-6
解得 x=5/2 y=5/6 此时最小值1/6琴生不等式 (x1^2+x2^2+...+xn^2)>=n[(x1+x2+...+xn)]^2 http://gzsx.cooco.net.cn/testdetail/181201/柯西不等式原式=46-30x+5x^2-20y+6xy+3y^2=5[x+(3/5)(y-5)]^2+(6/5)(y-5/6)^2+1/6当两个平方同时为零时,即x=(3/5)(5-y),y=5/6时,原式有最小值1/6
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