Find the minimum value of the constant term.  
Let $f(x)$ be a polynomial function with non-negative coefficients such that $f(1)=f’(1)=f’’(1)=f’’’(1)=1$. Find the minimum value of $f(0)$.
 
By Taylor’s formula, we can obtain
 
$$f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3+\cdots+\frac{f^{(n)}(1)}{n!}(x-1)^n.$$
 
Hence $$f(0)=\frac{1}{2}-\frac{1}{6}+\sum_{k=4}^n\frac{f^{(k)}(1)}{k!}(-1)^k.$$
 
This will help?

It is perhaps simpler to use Taylor's formula with fixed degree three and a Lagrange remainder:
$$
f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3+\frac{f^{(4)}(\xi)}{4!}(x-1)^4.
$$
where $\xi$ is between $1$ and $x$. For $x=0$ is $\xi \ge 0$ and the last term is non-negative, since $f^{(4)}$ has non-negative coefficients as well. This gives
$$
 f(0) \ge 1 - 1 + \frac 12 - \frac 16 = \frac 13 \, .
$$
The bound is sharp, equality holds for the function
$$
f(x)=1+(x-1)+\frac{1}{2!}(x-1)^2+\frac{1}{3!}(x-1)^3 = \frac 13 + \frac 12 x + \frac 16 x^3 \, .
$$
