不等式问题11已知x^2-2xy+2y^2

不等式问题11已知x^2-2xy+2y^2
不等式问题11
已知x^2-2xy+2y^2<=2
求证|x+y|<=√10

不等式问题11已知x^2-2xy+2y^2
x^2-2xy+2y^2<=2
===>
(x-y)^2+0.25y^2+0.25y^2+0.25y^2+0.25y^2<=2
根据算术平均数<=几何平均数,这个不等式
{(x-y+0.5y+0.5y+0.5y+0.5y)/5}<=
√{((x-y)^2+0.25y^2+0.25y^2+0.25y^2+0.25y^2)/5}=√(2/5)
==>(x+y)^2 <=10
==>|x+y|<=√10

(x+y)^2=x^2+2xy+y^2=5(x^2-2xy+2y^2)-(4x^2-12xy+9y^2)
=5(x^2-2xy+2y^2)+[-(2x-3y)^2]
因为 5(x^2-2xy+2y^2)<=10
-(2x-3y)^2<=0
所以 (x+y)^2<=10
所以 |x+y|<=√10