INMO 1998 question

Question

This was question 5 of 1998 INMO.

Suppose $a,b,c$ are three real numbers such that the quadratic equation $$ x^2 - (a +b +c )x + (ab +bc +ca) = 0 $$has roots of the form $\alpha + i \beta$ where $\alpha > 0$ and $\beta \not= 0$ are real numbers. Show that

(i) The numbers $a,b,c$ are all positive.

(ii) The numbers $\sqrt{a}, \sqrt{b} , \sqrt{c}$ form the sides of a triangle.

The only thing that I noticed about it was that if $P(x)= x^2 - (a +b +c )x + (ab +bc +ca)$, then $x P(x)=(x-a)(x-b)(x-c)+abc$, so then I saw the solution and have some questions about it.

They showed that $abc \neq 0$ and then said:

"If $abc < 0$ with, wlog, $a > 0$ then for $x$ close enough to $a$, $x P(x) < 0$, not possible.

$P(0) > 0$ and $a+b+c = 2 \alpha$

Every elementary cyclic polynomial of $a,b,c$ are positive so are they (classical by looking polynomial $(x-a)(x-b)(x-c)$)

$\Delta = \sum (a-b)^2 - \sum a^2 < 0$ f "

He said that his post had been truncated so, this might just be the proof of the first point.

I don't understand why he says that when $x$ is close enough to $a$, $x P(x) < 0$, is not possible. Wouldn´t it approach $a P(a) =abc < 0$

Also, shouldn't $abc>0$ be considered. I think that they are doing that to prove that $a,b$ and $c$ have to be positive, but I do not even know how to do that.

And what does the final summation mean? I know that it is related to cyclic polynomials but I do not know many things about them. I would like you to tell me where to find some stuff related to that if possible please.

I do know how to do the second part but it's ok if you tell me other ways to do it.

Answer 1

Because $$(a+b+c)^2-4(ab+ac+bc)<0$$ gives $$2(ab+ac+bc)-a^2-b^2-c^2>0.$$

Let it be not so. Id est, there is non-positive variable.

Since this inequality is not changed if substitute $a$ by $-a$, $b$ by $-b$ and $c$ by $-c$,

we can assume that $c\leq0$, $a\geq0$ and $b\geq0$ and after replacing $c$ by $-c$ we obtain: $$-2ac-2bc+2ab-a^2-b^2-c^2>0$$ or $$-c(c+2a+2b)-(a-b)^2>0$$ which is a contradiction.

Thus, we proved that $a$, $b$ and $c$ have the same sign and since by the given $a+b+c>0$,

we obtain that $a$, $b$ and $c$ are positives.

Now, we see that $$\sum_{cyc}(2ab-a^2)=(\sqrt{a}+\sqrt{b}+\sqrt{c})\prod_{cyc}(\sqrt{a}+\sqrt{b}-\sqrt{c})>0.$$ If $$\sqrt{a}+\sqrt{b}-\sqrt{c}<0$$ and $$\sqrt{a}-\sqrt{b}+\sqrt{c}<0$$ then after summing we obtain: $\sqrt{a}<0$, which is a contradiction.

Id est, $$\sqrt{a}+\sqrt{b}-\sqrt{c}>0,$$ $$\sqrt{a}-\sqrt{b}+\sqrt{c}>0,$$

$$\sqrt{b}+\sqrt{c}-\sqrt{a}>0$$ and we are done!

Answer 2

"If $abc<0$ with, wlog, $a>0$ then for $x$ close enough to $a$, $xP(x)<0$, not possible."

"$x$ close enough to $a$" implies $x>0$, which in turn implies $xP(x)<0 \Rightarrow P(x)<0$. But $P(x)>0$, because the parabola opens up and it has complex roots (meaning the parabola does not cross the $x$-axis.

Also, shouldn't $abc>0$ be considered.

It is similar logic. $abc>0$ implies all three positive (but then we are done) or two are negative. Wlog $a<0$ then for $x$ close enough to $a$, $xP(x)>0$, not possible, because $xP(x)>0 \Rightarrow P(x)<0$.