$b(a+c) = b(1-b) = b-b^2$ $a+c=1-b$ $2\sqrt{ac}\leq 1-b$ $ \sqrt{ac} \leq \frac{1-b}{2}$ $2ac \leq \frac{(1-b)^2}{2}$ $ab+bc+2ac \leq b(1-b)+ \frac{(1-b)^2}{2} =\frac{(1-b)(b+1)}{2} \leq \frac{1}{2}$ $(a,b,c) = (\frac{1}{2},0,\frac{1}{2})$