prove that in an acute angled triangle,cosAcosBcosC<=1/8

This can be proved by AM-GM inequality.

Let A,B,C be three acute angles of the triangle, So definitely Cos A , Cos B , Cos C

According to AM GM inequlaities

(Cos A + Cos B + Cos C)/3 >= (CosA CosB Cos C)^1/3

With the equality existing only when all terms are equal ie all the angle are 60 and Cos A= Cos B= Cos C = 1/2.

(CosA CosB Cos C)^1/3 <= (Cos A + Cos B + Cos C)/3

(CosA CosB Cos C)^1/3 <= (1/2 + 1/2 +1/2)/3

(CosA CosB Cos C)^1/3 <= (3/2)/3

(CosA CosB Cos C) <=(1/2)^3

(CosA CosB Cos C) <=1/8