F( 1)=a+b+c=0, so: C =-A-B a-b.

⊿=b^2-4ac=b^2+4a(a+b)=(2a+b)^2≥0

So: the image of f(x) intersects the x axis.

(2)g(x)=ax+b

F( 1)=a+b+c=0, a>b>c, so a>0.

Let t (x) = f (x)-g (x) = ax 2+(b-a) x+(c-b) = ax 2+(b-a) x+(-a-2b).

It is proved that when x≤-√3, t (x) >; 0 is constant.

T(x) symmetry axis x = (a-b)/2a >0, and the functional opening is upward.

So as long as t (-√ 3) >; 0, card.

t(-√3)= 3a+√3(a-b)-(a+2b)=(2+√3)(a-b)>0

Therefore, it is proved that.

(3) If x 1 x2 belongs to R and x 1

Let t (x) = f (x)-{f (x1)+f (x2)}/2.

Then: t (x1) = {f (x1)-f (x2)}/2; t(x2)= {f(x2)-f(x 1)}/2

t(x 1)*t(x2)=-{f(x 1)-f(x2)}^2/4≤0

So the solution of t(x)=0 must have a root between x 1 and x2.

That is, f(x)=f(x 1)+f(x2) must have a real root belonging to [x 1, x2].