Indija TS 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 $$\begin{array}{rl}f(-a)+f(-b)& =(b-a)(c-a)+(a-b)(d-b)=(a-b)(-c+a+d-b),\\ f(-c)+f(-d)& =(a-c)(d-c)+(b-d)(c-d)=(c-d)(-a+c+b-d).\end{array}$$ 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.