证明：若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2

证明：若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2
证明：若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2

证明：若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2
[g(x1)+g(x2)]/2=[x1²+x2²+a(x1+x2)+2b]/2=(x1²+x2²)/2+a(x1+x2)/2+b
g[(x1+x2)/2]=(x1+x2)²/4+a(x1+x2)/2+b
∴g[(x1+x2)/2]-[g(x1)+g(x2)]/2
=(x1+x2)²/4-(x1²+x2²)/2
=(x1²+2x1x2+x2²-2x1²-2x2²)/4
=-(x1-x2)²/4≤0
∴g[(X1+X2)/2]≤[g(x1)+g(x2)]/2