因为,
g( (x1+x2)/2 )= (1/4)(x1+x2)² + (a/2)(x1+x2) + b ;
(1/2)[g(x1)+g(x2)] = (1/2)(x1²+x2²) + (a/2)(x1+x2) + b ;
可得:
(1/2)[g(x1)+g(x2)] - g( (x1+x2)/2 )
= (1/2)(x1²+x2²) - (1/4)(x1+x2)²
= (1/4)(x1-x2)²
≥ 0 ,
所以,
g( (x1+x2)/2 )≤ (1/2)[g(x1)+g(x2)] .
因为,
g( (x1+x2)/2 )= (1/4)(x1+x2)² + (a/2)(x1+x2) + b ;
(1/2)[g(x1)+g(x2)] = (1/2)(x1²+x2²) + (a/2)(x1+x2) + b ;
可得:
(1/2)[g(x1)+g(x2)] - g( (x1+x2)/2 )
= (1/2)(x1²+x2²) - (1/4)(x1+x2)²
= (1/4)(x1-x2)²
≥ 0 ,
所以,
g( (x1+x2)/2 )≤ (1/2)[g(x1)+g(x2)] .