If in a triangle ABC, cosA cosB + sinA sinB sinC= 1, then a:b:c is

$\begin{array}{l}\cos A\cos B+\sin A\sin B\sin C=1\\ \Rightarrow \sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\le 1\\ \Rightarrow \cos A\cos B+\sin A\sin B\ge 1\\ \Rightarrow \cos (A-B)\ge 1\Rightarrow \cos (A-B)=1\Rightarrow A=B\\ \begin{array}{rl}\therefore & \sin C=\frac{1-{\cos }^{2}A}{{\sin }^{2}A}=1\Rightarrow C=\frac{\pi }{2}\\ \therefore & A=B=\frac{\pi }{4}\end{array}\\ \begin{array}{rl}\therefore & a:b:c=\sin A:\sin B:\sin C\\ & =1:1:\sqrt{2}\end{array}\end{array}$