<s>[INST] 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.
 [/INST] Observe that the equality fails if one of the $\cos$ is negative. We can assume that triangle is acute. By Jensen and followed by AM-GM inequality:
$x = \cos A, y = \cos B, z = \cos C\to x+y+z \leq \dfrac{3}{2}$:
$1 = 2xyz + xy+yz+zx \leq 2\cdot \dfrac{(x+y+z)^3}{27}+\dfrac{(x+y+z)^2}{3}\leq \dfrac{2}{27}\cdot \left(\dfrac{3}{2}\right)^3+\dfrac{1}{3}\cdot \left(\dfrac{3}{2}\right)^2=1\to $ equality must occur when $x = y = z \to $ triangle is equilateral.
</s>