, 2008

We have to consider two possibilities: $x_1 + x_2 \le -1$ or $x_1 + x_3 \le -1$. (Both the roots are from the same equation or each root coming from a different equation.) Suppose $x_1 + x_2 \le -1$. Then $-a = x_1 + x_2$ shows that $a > 0$. Since the equations $P(x) = \alpha$ and $P(x) = \beta$ have distinct real roots, we have
$$4(b - \alpha) < a^2, \quad 4(b - \beta) < a^2.$$
Thus $4b < a^2 + 2(\alpha + \beta) = a^2 - 2a = a(a - 2) < 0$. Hence $b < 0$ in this case.
Suppose $x_1 + x_3 \le -1$. We have $P(x_1) = \alpha$, $P(x_3) = \beta$, so that
$$x_1^2 + x_3^2 + a(x_1 + x_3) + 2b = \alpha + \beta = -a.$$
If $a > 0$, we see that
$$(x_1 + \frac{a}{2})^2 + (x_3 + \frac{a}{2})^2 + 2b = -a + \frac{a^2}{2} = \frac{a(a-2)}{2},$$
so that $2b \le a(a - 2)/2 < 0$. If $a \le 0$, we have
$$x_1^2 + x_3^2 + 2b = -a(x_1 + x_3 + 1) \le 0,$$
so that $b \le 0$. We conclude that $b \le 0$ in all cases.
Now it is easy to see that
$$P(x+y) = P(x) + P(y) + 2xy - b \ge P(x) + P(y),$$
for all non-negative $x, y$.