India — Team Selection Test, 2012

Let $f(x) = (x + a)(x + d) + (x + b)(x + c)$. The leading coefficient of $f(x)$ is $2$. Hence $f(x)$ is positive for large values of $x$. We have
$$\begin{aligned} f(-a) + f(-b) &= (b-a)(c-a) + (a-b)(d-b) = (a-b)(-c+a+d-b), \\ f(-c) + f(-d) &= (a-c)(d-c) + (b-d)(c-d) = (c-d)(-a+c+b-d). \end{aligned}$$
We know that $a \ge b$ and $c \ge d$. If $f(-a) + f(-b) \le 0$, then we are done; then either $f(-a) \le 0$ or $f(-b) \le 0$ so that $f(x)$ crosses the $x$-axis at some point. Otherwise $-c+a+d-b > 0$. This implies that $-a+c+b-d < 0$. But then
$$f(-c) + f(-d) = (c - d)(-a + c + b - d) \le 0.$$
We conclude that either $f(-c) \le 0$ or $f(-d) \le 0$. Again $f(x)$ crosses the $x$-axis somewhere.