Vietnam Mathematical Olympiad

Denote by $\alpha$ and $\beta$ the common differences of the arithmetic progressions $a_n$ and $b_n$, respectively. The quadratics $P_k$ ($k = 1, 2, \dots, m$) can be written as
$$P_k(x) = x^2 + (a_1 + (k-1)\alpha)x + b_1 + (k-1)\beta, \quad k = 1, 2, \dots, m.$$
Since $P_1$ and $P_m$ have no real root, we have
$$\Delta_1 = a_1^2 - 4b_1 < 0$$
and
$$\Delta_m = (a_1 + (m-1)\alpha)^2 - 4(b_1 + (m-1)\beta) < 0.$$
Suppose that there exists $k \le m$ such that the quadratic $P_k$ has a real root. We have
$$\Delta_k = (a_1 + (k-1)\alpha)^2 - 4(b_1 + (k-1)\beta) \ge 0.$$
Since $k, m > 1$, it follows from the above and by the previous inequalities that
$$(k-1)\Delta_m < 0 < (m-1)\Delta_k,$$
which is equivalent to
$$(k-1)a_1^2 + (k-1)(m-1)^2\alpha^2 - 4(k-1)b_1 < (m-1)a_1^2 + (m-1)(k-1)^2\alpha^2 - 4(m-1)b_1.$$
This implies that
$$\begin{align*} (k-1)(m-1)(m-k)^2\alpha^2 &< (m-k)a_1^2 - 4(m-k)b_1 \\ (k-1)(m-1)^2\alpha^2 &< a_1^2 - 4b_1 \quad (\text{since } m > k) \\ \Rightarrow a_1^2 - 4b_1 &\ge 0, \end{align*}$$
which contradicts $\Delta_1 = a_1^2 - 4b_1 < 0$. This completes our proof.