An algebra problem by Anoopam Mishra

We will find the maximum and minimum values separately.

$\textbf{Minimum}$ : Since $a,b$ and $c$ are real numbers, then

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=1+2(ab+bc+ac)\ge0$

$\Rightarrow ab+bc+ac\ge-\frac{1}{2}$

for which equality can be attained when $a,b$ and $c$ satisfy the system

$\begin{cases} a^2+b^2+c^2=1\\ a+b+c=0\\ ab+bc+ac=-\frac{1}{2}\\ \end{cases}$

One such triplet is $(a,b,c)=(\frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}})$ .

$\textbf{Maximum}$ : We prove the well known inequality

$a^2+b^2+c^2\ge ab+ac+bc$

for real variables. We consider the two sequences $a,b,c$ and $a,b,c$ . Notice that in $a^2+b^2+c^2$ , the largest two terms from each of the sequences is paired off as so is the other two terms in order. Thus by the Rearrangement Inequality, $a^2+b^2+c^2$ is the largest possible pairing of each term from one sequence to the other, in particular,

$a^2+b^2+c^2=1\ge ab+bc+ac$

One way equality is achieved is when $a=b=c=\frac{1}{\sqrt{3}}$

(AM-GM might've not worked in proving the above inequality due to fact that the variables are not restricted to being nonnegative.)

Combining the arguments above gives that

$ab+bc+ac\in\boxed{[-0.5,1]}$

we can use cauchy scwarz in equality