Let $c_j = x - b_j > 0$ for each $j$. Note that
$$f(y) = (y - b_1)(y - b_2)\cdots(y - b_n).$$
Therefore, we have
$$\begin{aligned} f(x+1) &= (x+1-b_1)(x+1-b_2)\cdots(x+1-b_n) \\ &= (1+c_1)(1+c_2)\cdots(1+c_n). \end{aligned}$$
Now, by the AM-GM inequality, we have
$$\frac{1}{c_1} + \frac{1}{c_2} + \cdots + \frac{1}{c_n} \ge n \sqrt[n]{\frac{1}{c_1 c_2 \cdots c_n}}.$$
Thus, it suffices to prove
$$\prod_{j=1}^{n} \sqrt[\frac{1}{n}]{\frac{(1+c_j)^n}{c_j}} \ge 2n. \quad (1)$$
Indeed, since $n \ge 2$, we get
$$\frac{(1+c_j)^n}{c_j} \ge \frac{1}{c_j} + n + \frac{n(n-1)}{2} c_j \ge n + 2 \sqrt{\frac{1}{c_j} \cdot \frac{n(n-1)}{2}} c_j \ge 2n$$
by the AM-GM inequality and a simple inequality $2\sqrt{\frac{n(n-1)}{2}} \ge n$ (which is equivalent to $n \ge 2$). This implies (1). Equality holds when $n=2$ and $c_j = 1$, i.e. $x-1 = b_1 = b_2 = \cdots = b_n$.