Let f(x) = x^3+ ax^2+ bx + c with a, b, c real. Show that

M = max|f(x)|≥1/4
-1 ≤ x ≤ 1
and find all cases where equality occurs.

so far i have

Let h(x)=x^3, then let g(x)=mx, we need to find m so that g(x) can get as close to h(X) with an even minimum distance between them, between [-1,1]
By trail and error we get m=3/4and are left with
f(x)=x^3-3/4x where a=0, b=-3/4 and c=0
So 1,-1 and x_2 are the critical points. So when is g(x)-h(x)=1/4?
3/4x-x^3=1/4
x^3-3/4x+1/4=0
4x^3+3x+1=0
So the only solution are when x=0.5 or -1 and we have
f(0.5)=0.125-0.385=-0.26 so |f(0.5)|=0.26>1/4
f(-1)=-1+3/4=-0.25 so |f(-1)|=0.25=1/4
So when x=0.5 or -1, m>=1/4.

but apparently this doesnt make much sense,
any help would be much appreciated.