Question

Let f : R → R be the function
f(x) = (x – a1)(x – a2) + (x – a2)(x – a3) + (x – a3)(x – a1)
with a1, a2, a3 ∈ ı . Then f(x) ≥ 0 if and only if :

Answer 1

Question #36597, Calculus

Let $f: \mathbb{R} \to \mathbb{R}$ be the function

$f(x) = (x - a1)(x - a2) + (x - a2)(x - a3) + (x - a3)(x - a1)$ with $a1, a2, a3 \in \mathbb{R}$. Then $f(x) \geq 0$ if and only if:

Solution

Open brackets in the expression of $f(x)$, then

$$\begin{array}{l} f(x) = x^2 - (a_1 + a_2)x + a_1a_2 + x^2 - (a_2 + a_3)x + a_2a_3 + x^2 - (a_1 + a_3)x + a_1a_3 = \\ = 3x^2 - 2(a_1 + a_2 + a_3)x + (a_1a_2 + a_2a_3 + a_1a_3). \end{array}$$

Take into account that this parabola opens upward.

Moreover, $f(x) \geq 0$ if and only if

$$\begin{array}{l} D = 4(a_1 + a_2 + a_3)^2 - 4 * 3 * (a_1a_2 + a_2a_3 + a_1a_3) = \\ = 4(a_1^2 + a_2^2 + a_3^2 + 2a_1a_2 + 2a_2a_3 + 2a_1a_3) - 12a_1a_2 - 12a_2a_3 - 12a_1a_3 = \\ = 4(a_1^2 + a_2^2 + a_3^2 - a_1a_2 - a_2a_3 - a_1a_3) = \\ = 4 * \left(\frac{a_1}{\sqrt{2}} - \frac{a_2}{\sqrt{2}}\right)^2 + 4 * \left(\frac{a_2}{\sqrt{2}} - \frac{a_3}{\sqrt{2}}\right)^2 + 4 * \left(\frac{a_1}{\sqrt{2}} - \frac{a_2}{\sqrt{2}}\right)^2 \leq 0. \end{array}$$

But we know that the sum of squares is non-negative.

Therefore, the only possible variant is

$$\frac{a_1}{\sqrt{2}} - \frac{a_2}{\sqrt{2}} = 0, \quad \frac{a_2}{\sqrt{2}} - \frac{a_3}{\sqrt{2}} = 0, \quad \frac{a_1}{\sqrt{2}} - \frac{a_3}{\sqrt{2}} = 0,$$

from where we conclude

$$a_1 = a_2 = a_3.$$

Answer: $a_1 = a_2 = a_3$