An algebra problem by Ossama Ismail

$\begin{aligned}(a+b+c)^2&=a^2+b^2+c^2+2(ab+ac+bc)\\ &=1+2(ab+ab+bc)\\ ab+ab+bc&=\frac{(a+b+c)^2-1}{2}\end{aligned}$ The minimum value of anything squared is 0, so $ab+ac+bc$ is minimised when $(a+b+c)=0$ .

If we let $c=0$ , an intuitive example of this case emerges: $\begin{aligned}a^2+b^2+0^2&=1&a+b+0&=0\\ a^2+b^2&=1&a+b&=0\\ &&b&=-a\\ a^2+(-a)^2&=1&\longleftarrow\\ 2a^2&=1\\ a&=\frac{1}{\sqrt2}\\ &&\implies b&=-\frac{1}{\sqrt2}\end{aligned}$ $a=\frac{1}{\sqrt2}$ , $b=-\frac{1}{\sqrt2}$ and $c=0$ fits the original question, as $\left(\frac{1}{\sqrt2}\right)^2+\left(-\frac{1}{\sqrt2}\right)^2+0^2=1$ .

Now we can substitute our values of $a$ , $b$ and $c$ into our equation for $ab+ac+bc$ knowing that they will give the minimum value: $\begin{aligned}ab+ab+bc&=\frac{(a+b+c)^2-1}{2}\\ &=\frac{(\frac{1}{\sqrt2}-\frac{1}{\sqrt2}+0)^2-1}{2}\\ &=\frac{(0)^2-1}{2}\\ &=-\frac{1}{2}\end{aligned}$