题目

已知f(x )=x^2+ax+b,p+q=1,0

提问时间：2021-01-14

答案

pf(x)+qf(y)>=f(px+qy)
<=> px^2+pax+pb+qy^2+qay+qb>=(px+qy)^2+apx+aqy+b
<=> px^2+qy^2>=(px+qy)^2
<=> px^2+qy^2>=p^2x^2+q^2y^2+2pqxy
<=> (p-p^2)x^2+(q-q^2)y^2>=2pqxy
将q=1-p代入,化简得
(p-p^2)(x^2+y^2)>=2(p-p^2)xy
∵ x^2+y^2>=2xy
∴ p-p^2>0
<=> p>p^2
<=> 0<=p<=1