Indija TS 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 $$4(b-\alpha )<a^2,4(b-\beta )<a^2.$$ 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 $x,y$.