高1数学，数学题，高手进，急~

(1) 若f(x)=ax+b,则f(X1+X2/2)=f(X1)+f(X2)/2
f(x1+x2/2)=a(x1+x2)/2+b
f(x1)+f(x2)=ax1+b+ax2+b=a(x1+x2)+2b
f(x1)+f(x2)/2=a(x1+x2)/2+b
所以f(X1+X2/2)=f(X1)+f(X2)/2

(2) 若g(x)=X2+ax+b,则g(x1+x2/2) <= g(x1)+g(x2)/2
g(x1+x2/2)=(x1+x2)^2/4+a(x1+x2)/2+b
[g(x1)+g(x2)]/2=[x1^2+x2^2+a(x1+x2)+2b]/2
g(x1+x2/2) - [g(x1)+g(x2)]/2
=(x1+x2)^2/4-(x1^2+x2^2)/2=-(x1-x2)^2/4≤0
所以g(x1+x2/2) ≤ g(x1)+g(x2)/2

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f((x1+x2)/2)=a[(x1+x2)/2]+b=ax1/2+ax2/2+b/2+b/2=(ax1+b)/2+(ax2+b)/2=f(x1)/2+f(x2)/2=(f(x1)+f(x2))/2
g((x1+x2)/2)=(x1+x2)^4/4+a(x1+x2)/2+b
(g(x1)+g(x2))/2=(x1^2+x2^2)/2+a/2(x1+x2)+b
因为(x1-x2)^2≥0，所以x1^2+x2^2≥2x1x2,不等式两边同时加x1^2+x2^2得2(x1^2+x^2)≥(x1+x2)^2,不等式两边同除以4得(x1^2+x2^2)/2≥(x1+x2)^2/4
所以(x1^2+x2^2)/2+a/2(x1+x2)+b≥(x1+x2)^4/4+a(x1+x2)/2+b
即g((x1+x2)/2)≤(f(x1)+f(x2))/2