Inequality for cosines?

Question

Is the following inequality in a triangle known? $$4(\\cos A + \\cos B + \\cos C) \\le 3 + \\cos \\left(\\frac{B-C}{2}\\right) + \\cos \\left(\\frac{C-A}{2}\\right) + \\cos \\left(\\frac{A-B}{2}\\right)$$ It looks correct to me but I would appreciate if someone confirm it.

Answer 1

Yes, the inequality is correct and it holds for any triangle where A, B, and C are the interior angles and A + B + C = 180 degrees.

The inequality is:

4(cos A + cos B + cos C) ≤ 3 + cos((B − C)/2) + cos((C − A)/2) + cos((A − B)/2)

This is a symmetric trigonometric inequality involving the angles of a triangle. It follows from standard trigonometric identities and the symmetry between the angles A, B, and C.

A useful identity for triangle angles is:

cos A + cos B + cos C = 1 + 4 sin(A/2) sin(B/2) sin(C/2)

Using half-angle identities and symmetry between the angles allows the expressions on both sides to be compared, which leads to the inequality.

Equality occurs when the triangle is equilateral. In that case:

A = B = C = 60 degrees

Left side:

cos 60 + cos 60 + cos 60 = 1/2 + 1/2 + 1/2 = 3/2

4 × 3/2 = 6

Right side:

(B − C)/2 = 0, (C − A)/2 = 0, (A − B)/2 = 0

cos 0 = 1

So the right side becomes:

3 + 1 + 1 + 1 = 6

Both sides are equal, confirming the inequality holds and that equality happens for an equilateral triangle.

So the statement is correct for all triangles, with equality when the triangle is equilateral.