Vieta’s Formulas for Cubic Polynomial

Let $x_1$, $x_2$ and $x_3$ be the roots of the polynomial

$$ ax^3 + bx^2 + cx + d $$

Vieta’s Formulas state that

$$ x_1+x_2+x_3 = – \frac{b}{a} $$

$$ x_1x_2+x_2x_3+x_3x_1 = \frac{c}{a} $$

$$ x_1x_2x_3 = -\frac{d}{a} $$

Problem (Tournament of Towns, 1985)

Given the real numbers $a$, $b$, $c$, such that $a+b+c>0$, $ab+bc+ac>0$, $abc>0$. Prove that $a>0$, $b>0$ and $c>0$.

Solution

Let us consider a polynomial with the roots $x=a$, $x=b$ and $x=c$:

$$ f(x) = (x-a)(x-b)(x-c) $$

where

$$ f(a)=f(b)=f(c)=0 $$

Then by Vieta’s Formulas we have

$$ f(x) = x^3 -(a+b+c)x^2 + (ab+bc+ca)x -abc $$

If one of the numbers $a$, $b$ or $c$ is zero, then $abc=0$ and we have a contradiction with the condition that $abc>0$.

If one of the numbers $a$, $b$ or $c$ is negative, then for $x<0$ we have

$$x^3<0$$

$$-(a+b+c)x^2<0$$

$$(ab+bc+ca)x<0$$

$$-abc<0$$

This implies that

$$ f(x) = x^3 -(a+b+c)x^2 + (ab+bc+ca)x -abc < 0 $$

which contradicts the assumption that

$$ f(a)=f(b)=f(c)=0 $$

and that one of the numbers $a$, $b$ or $c$ is negative.

Therefore numbers $a$, $b$ or $c$ cannot be zero nor negative and thus are all positive numbers.

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