以p+q=1为条件
pf(x)+qf(y)-f(px+qy)=px^2+qy^2-(px+qy)^2(化简后)
=(1-p)x^2+(1-q)y^2-2pqxy 又p+q=1
=qx^2+py^2-2pqxy
=(px+qy)^2≥0
则pf(x)+qf(y)≥f(px+qy
以pf(x)+qf(y)≥f(px+qy)为条件
则移项合并得(1-p)x^2+(1-q)y^2-2pqxy≥0(化简后) 又0≤p≤1
(1-p)x^2+(1-q)y^2≥2根号下(1-p)(1-q)乘以xy ≥2pqxy
根号下(1-p)(1-q)≥pq
1-p-q≥0 1≥ p+q
则p+q=1时成立