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Let $f: R \rightarrow R$ be the function $f(x)=\left(x-a_1\right)\left(x-a_2\right)$ $+\left(x-a_2\right)\left(x-a_3\right)+\left(x-a_3\right)\left(x-a_1\right)$ with $a_1, a_2, a_3 \in R$.Then, $f(x) \geq 0$ if and only if
at least two of $a_1, a_2, a_3$ are equal
$a_1=a_2=a_3$
$a_1, a_2, a _3$ are all distinct
$a_1, a_2, a _3$ are all positive and distinct
Solution
(b)
We have,
$f(x)=\left(x-a_1\right)\left(x-a_2\right)+\left(x-a_2\right)\left(x-a_3\right)+\left(x- a _3\right)\left(x- a _1\right)$
$f(x)=3 x^2-2\left(a_1+a_2+a_3\right) x+a_1 a_2+a_2 a_3+a_1 a_3$
$f(x) \geq 0$ iff $D \leq 0$
$\therefore 4\left(a_1+a_2+a_3\right)^2-4 \times 3 \left(a_1 a_2+a_2 a_3+a_1 a_3\right) \leq 0$
$=a_1^2+a_2^2+a_3^2-\left(a_1 a_2+a_2 a_3+a_1 a_3\right) \leq 0$
$=\frac{1}{2}\left[\left(a_1-a_2\right)^2+\left(a_2-a_3\right)^2+\left(a_3-a_1\right)^2\right] \leq 0$
It is possible only
$a _{1 .}=a_2, a_2= a _3, a _3= a _1$
$\therefore \quad a_1=a_2=a_3$
