I am solving a math contest problem. Assume we have the quadratic equation $x^2+a_1x+a_2=0$ where $a_1,a_2\in \mathbb{R}$ are real numbers. The roots of this equation can be found as (from equation it can be inferred that $a_1 = -(p_1+p_2), a_2 = p_1p_2$)
$$p_1,p_2 = -\frac{a_1}{2}\pm \sqrt{\frac{a_1^2}{4}-a_2}$$
Now assume that $|p_1|< 1, |p_2|< 1$, in other words the module of the roots is lesser than unity (in the case of real roots, it translates to the fact that their absolute value is less than one). The claim is twofold
-
$|a_2| < 1$
-
$|a_1| < a_2+1$
proving the first one very easy since $|a_2|=|p_1p_2| \le |p_1||p_2|< 1$. However, the second one is almost impossible to prove! If we assume the roots are real, then by using $|p_i|< 1$ we can say
$$-1< -\frac{a_1}{2}\pm \sqrt{\frac{a_1^2}{4}-a_2}< 1 \Rightarrow -1+\frac{a_1}{2} < \pm \sqrt{\frac{a_1^2}{4}-a_2}< 1 + \frac{a_1}{2}$$ then this gives two inequalities which can be manipulated to get similar results with a lot of mental gymnastics! But I don't know what to do in general! Is there any simpler way to reach the conclusion? What can be done in general case (complex conjugate roots and real distinct roots)?
Thank you!
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For the complex conjugate pair, we can write $p_1 = p_r+ip_i, p_2 = p_1-ip_i$ and by substitution we get two true statements
$$(1+p_r)^2+p_i^2 > 0 , (p_r-1)^2+p_i^2 > 0$$
and this prove the complex case. However, I am trying to find a way to drive the second statement using the given assumptions! Is there a way?
Best Answer
By Vieta, $p_1+p_2=-a_1$ and $p_1p_2=a_2$
As you say, it's trivial to show that $\big|a_2\big|<1$.
Let's assume that $\big|a_1\big| \ge a_2+1$ and try to reach a contradiction. Substituting and squaring both sides, \begin{align}\left(p_1+p_2\right)^2 &\ge \left(p_1 p_2+1\right)^2 \\ p_1^2+2p_1 p_2 +p_2^2 &\ge p_1^2 p_2^2+2p_1 p_2+1 \\ p_1^2 +p_2^2 &\ge p_1^2 p_2^2+1\\ 0&\ge\left(p_1^2-1\right)\left(p_2^2-1\right) \end{align}
If $p_1$ and $p_2$ are real, this means either $p_1^2-1\le 0$ and $p_2^2-1\ge 0$, or vice-versa. But this immediately contradicts the condition that both $\big|p_1\big|<1$ and $\big|p_2\big|<1$.
So the only possibility is that $p_1$ and $p_2$ are a complex conjugate pair. But then $p_1^2-1$ and $p_2^2-1$ are also a conjugate pair, whose product must be positive; contradiction!
Therefore $\big|a_1\big| < a_2+1$.