I encountered the following nice little fact in a little problem of mine, and would like to know what proofs you could have in mind !
Consider $a_0, \ldots, a_n$ $n$ real numbers and form the polynomial:
$$P(x) =\sum_{i=0}^n a_i x^i.$$
Prove that one has the following bound on the value of the polynomial: $$\forall x \in [0,1], \min\{ \sum_{0 \leq i \leq j} a_i, j =0 \ldots n\} \leq P(x) \leq \max\{ \sum_{0 \leq i \leq j} a_i, j =0 \ldots n\}.$$
The question is for instance a useful addition to :
How to obtain uniform bounds on a polynomial by looking at its coefficients.
Best Answer
Let $q_j=\sum_{0 \leq k \leq j} a_k, j=0,..n$ and note that $P(0)=q_0, P(1)=q_n$ so to prove the inequality in the OP we can assume $x \in (0,1)$. Let $q,Q$ the min and max in cause so $q \le q_k \le Q, k=0,...n$
Now we note that $\frac{P(x)}{1-x}=\sum_{k=0}^nq_kx^k+\sum_{k \ge n+1}q_nx^k, 0<x<1$
In particular since $q \le q_k \le Q$ we have $$\frac{q}{1-x} \le \sum_{k=0}^nq_kx^k+\sum_{k \ge n+1}q_nx^k \le \frac{Q}{1-x}$$ hence $$\frac{q}{1-x} \le \frac{P(x)}{1-x} \le \frac{Q}{1-x}$$
Since $x \in (0,1)$ the required inequality $q \le P(x) \le Q$ follows.