If in a triangle $ABC$,$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A$,then prove that triangle will be equilateral triangle If in a triangle $ABC$ we have
$$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A\ ,$$
then the triangle will be equilateral triangle.
I tried but except few steps,could not prove it.  Putting $A=B=C=\frac{\pi}{3}$ makes both sides equal.  How should i prove it?Please help me.

For brevity write $a=\cos A$, $b=\cos B$, $c=\cos C$.  Then we have
$$C=\pi-A-B$$
and so
$$c=-\cos(A+B)=-ab+\sin A\sin B$$
and so
$$(c+ab)^2=(1-a^2)(1-b^2)\ .$$
Expanding, rearranging and using the identity you are given,
$$a^2+b^2+c^2=1-2abc=ab+bc+ca\ .$$
Now
$$(a-b)^2+(b-c)^2+(c-a)^2=2a^2+2b^2+2c^2-2ab-2bc-2ca$$
and so
$$(a-b)^2+(b-c)^2+(c-a)^2=0\ ,$$
which shows that $a,b,c$ are all equal; hence $A,B,C$ are all equal.
