Reimann Hypothesis Forum

Trigonometric Substitutions & Convexity in High School Olympiad Inequalities

Whenever a symmetric three-variable inequality comes with the constraint ab + bc + ca = 1 or a + b + c = abc, substituting a = tan(α/2) or a = tan A turns algebraic radicals into clean triangle identities.

Agreed. Mixing Jensen's inequality on f(x) = sec^2(x) or tan(x) on (0, π/2) with acute-triangle constraints often avoids messy Lagrange multipliers altogether.