How to Solve Tricky ITF Inequalities Using Titu's Lemma | JEE #jee

In this video, we solve a challenging Inverse Trigonometric Functions (ITF) inequality problem for JEE Main & JEE Advanced. We use substitution along with Titu's Lemma (Cauchy-Schwarz Inequality) and discuss why equality does not hold based on the domain/range of tan⁻¹(x). 📌 Problem Statement: If a, b, c are positives and a + b + c = abc, prove that: 1/tan⁻¹(a) + 4/tan⁻¹(b) + 9/tan⁻¹(c) greater than 36/π ⏱️ Timestamps: 00:00 - Problem Introduction & Tan⁻¹(x) Graph 01:34 - Substitution: x = tan⁻¹(a), y = tan⁻¹(b), z = tan⁻¹(c) 02:00 - Using Property: tan(x) + tan(y) + tan(z) = tan(x)tan(y)tan(z) 03:34 - Applying Titu's Lemma to Prove the Inequality 05:15 - Condition for Equality & Finding Constants 07:33 - Why Strict Inequality Holds (Graph Range Analysis) Tags (comma-separated for Studio): ITF question, inverse trigonometric functions, titus lemma, cauchy schwarz inequality, jee advanced maths, jee maths, jee advanced, itf inequality, inverse trigonometry questions, siyesh iit jee, iit jee mathematics, jee 2026, jee preparation Don't forget to Like, Share, and Subscribe to @Siyesh_IIT-JEE for more advanced mathematics problem solutions! #JEE #JEEAdvanced #JEEMaths #InverseTrigonometricFunctions #IITJEE