$\displaystyle f(0)=f(1)=0, f''(t)\geq0,\Rightarrow f(t)\leq0, t\in[0,1] $
prove this claim
Assume there is a c in (0,1) such that f(c)>0, then (by the mean value theorem) there are points u in (0,c) and v in (c,1) such that f'(u)>0 and f'(v)<0.
But f''(x)>=0, x in [0,1] implies that f'(x) is non-decreasing on [0,1], which contradicts the above.
Hence there is no point c in (0,1) such that f(c)>0 which implies that f(x)<=0 for all x in [0,1].