1. **Express $P(1)$ and $P(1)^2$:**
$$P(1) = \sum_{i=0}^{n} a_i$$
$$P(1)^2 = \left( \sum_{i=0}^{n} a_i \right)^2$$

2. **Expand $P(1)^2$ using the binomial theorem:**
$$P(1)^2 = \left( \sum_{i=0}^{n} a_i \right)^2 = \sum_{i=0}^{n} a_i^2 + 2 \sum_{0 \leq i < j \leq n} a_i a_j$$

3. **Express $P(X)^2$ in terms of its coefficients:**
$$P(X)^2 = b_{2n} X^{2n} + b_{2n-1} X^{2n-1} + \cdots + b_{n+1} X^{n+1} + \cdots + b_1 X + b_0$$

4. **Identify the coefficient $b_{n+1}$ in $P(X)^2$:**
The coefficient $b_{n+1}$ is the sum of products of coefficients $a_i$ and $a_j$ such that $i + j = n+1$:
$$b_{n+1} = \sum_{i=0}^{n} a_i a_{n+1-i}$$

5. **Relate $b_{n+1}$ to $P(1)^2$:**
Since $0 \leq a_i \leq a_0$ for all $i$, we have:
$$b_{n+1} = \sum_{i=0}^{n} a_i a_{n+1-i} \leq \sum_{i=0}^{n} a_i a_0 = a_0 \sum_{i=0}^{n} a_i$$

6. **Use the inequality to bound $4b_{n+1}$:**
$$4b_{n+1} \leq 4 a_0 \sum_{i=0}^{n} a_i$$

7. **Compare with $P(1)^2$:**
$$P(1)^2 = \left( \sum_{i=0}^{n} a_i \right)^2$$
Since $a_0 \leq \sum_{i=0}^{n} a_i$, we have:
$$4 a_0 \sum_{i=0}^{n} a_i \leq 4 \left( \sum_{i=0}^{n} a_i \right)^2$$

8. Conclude the inequality:
$$4b_{n+1} \leq 4 \left( \sum_{i=0}^{n} a_i \right)^2 = P(1)^2$$

$\blacksquare$