x2+y2+z2=1求 x+y取值范围_数学解答

x2+y2+z2=1求 x+y取值范围

人气：237 ℃　时间：2020-04-07 10:36:05

解答

x^2+y^2+z^2=1
=>x^2+y^2=1-z^2
=>(x+y)^2-2xy=1-z^2
(x+y)^2=x^2+y^2+2xy>=4xy
=>xy=(x+y)^2-(x+y)^2/4*2=(x+y)^2/2
=>(x+y)^2