是不是要证明?√x1^2+x2^2+√y1^2+y2^2大于等于√[(x1+y1)^2+(x2+y2)^2]
只需证明(√x1^2+x2^2+√y1^2+y2^2)^2-[(x1+y1)^2+(x2+y2)^2]大于等于0
(√x1^2+x2^2+√y1^2+y2^2)^2-[(x1+y1)^2+(x2+y2)^2]
=x1^2+x2^2+y1^2+y2^2+2√x1^2+x2^2*√y1^2+y2^2-[x1^2+2x1y1+y1^2+x2^2+2x2y2+y2^2]
=2√x1^2+x2^2*√y1^2+y2^2-2x1y1-2x2y2
上式大于等于0 只需证明
【2√x1^2+x2^2*√y1^2+y2^2】^2大于等于(2x1y1+2x2y2)^2
【2√x1^2+x2^2*√y1^2+y2^2】^2-(2x1y1+2x2y2)^2
=4(x1^2+x2^2)(y1^2+y2^2)-4(x1y1+x2y2)^2
=4[x1^2Y1^2+X2^2y2^2+x1^2y2^2+x2^2y1^2]-4(x1^2Y1^2+X2^2y2^2+2x1x2y1y2)
=4(x1y2-x2y1)^2≥0
得证