A(x1,y1)
B(x2,y2)
M点((x1+x2)/2,(y1+y2)/2)
(x1+x2)/2*(y1+y2)/2=0
x1y1+x2y1+x1y2+x2y2=0
2x1=y1
2x2=y2
所以
3x1^2+2x1x2+2x1x2+2x2^2=0
3x1^2+4x1x2+2x2^2=0
(x1+x2)/2>=√(x1x2)
A(x1,y1)
B(x2,y2)
M点((x1+x2)/2,(y1+y2)/2)
(x1+x2)/2*(y1+y2)/2=0
x1y1+x2y1+x1y2+x2y2=0
2x1=y1
2x2=y2
所以
3x1^2+2x1x2+2x1x2+2x2^2=0
3x1^2+4x1x2+2x2^2=0
(x1+x2)/2>=√(x1x2)