We have 1-8.cos(A).cos(B).cos(C) = 1-4.cos(A)(cos(B-C)+cos(B+C)).Since cos(B+C)= -cos(π-B-C)= -cos(A), the expression above is: 1-4.cos(A).cos(B-C) + 4.cos(A)² = sin(B-C)² + cos(B-C)² -4.cos(A).cos(B-C) + 4.cos(A)² = sin(B-C)² + [cos(B-C)-2cos(A)]² which is always non negative. |