Question

Let $f:R\to R$ be the function $f\left(x\right)=(x-a_1)(x-a_2)+(x-a_2)(x-a_3)+(x-x_3)(x-x_1)$ with $a_1,a_2,a_3\in R$ Then f(x)$\ge 0$ if and only if

• A. at least two of $a_1,a_2,a_3$ are equal
• B. $a_1=a_2=a_3$
• C. $a_1,a_2,a_3$ are all distinct
• D. $a_1,a_2,a_3$ are all positive and3 distinct

Answer 1

The correct option is B $a_1=a_2=a_3$

$f\left(x\right)=(x-a_1)(x-a_2)+(x-a_2)(x-a_3)+(x-a_3)(x-a_1)$

$f\left(x\right)=3x^2-2(a_1+a_2+a_3)x+(a_1{a}_2+a_2{a}_3+a_3{a}_1)$

Given $f\left(x\right)\ge 0$

$\therefore 4{(a_1+a_2+a_3)}^2-12(a_1{a}_2+a_2{a}_3+a_3{a}_1)\le 0$

$4a_1^2+4a_2^2+4a_3^2-4a_2{a}_3-4a_3{a}_1-4a_1{a}_2\le 0$

${(a_1-a_2)}^2+{(a_2-a_3)}^2+{(a_3-a_1)}^2$

$\therefore$Sum must be equal to zero

${(a_1-a_2)}^2+{(a_2-a_3)}^2+{(a_3-a_1)}^2=0$ only if $a_1=a_2=a_3$