The Area Way Incases , where a relation with areas is possible, It is quite efficient to use it to solve the problem. Example – If are the altitudes of a triangle from A,B,C to the opposite sides and r is the inradius. Prove that – 1. 2. Proof Let I be the incentre. Note […]
With the on-going trend, Algebraic Inequalities are almost erasing the existence of Geometrical Inequalities . Nevertheless, They are equally beautiful (if not more) , although they seem a lot harder to me. Yes, they can mostly be proved by brute-force trigononmetry or complex (whatever it is , they are equally good solutions) , but I […]
Problem Prove that – P.S. It is straightforward holder, but what about a solution by Cauchy or AM-GM ? Solution By, AM-GM we have , Write up analogous inequalities, and add to get – which re-arranges to the desired question 🙂
Problem Prove that for positive reals with sum 1 , Proof Homogenize the inequality to – Since the inequality is cyclic, let Substitute – and let , The inequality by the above substitution is transformed to – By AM-GM inequality , therefore the inequality is true with equality which gives us the equality cases –
Problem Let be nonnegative numbers, no two of which are zero. Prove that: Proof It is equivalent to – Therefore , Let , So, and also note that – thus the inequality is proved.
Problem For positive reals prove that – Proof Let then the inequality is equivalent to – Obviously we have that – (since ) Also by AM-GM , multiplying these two we get the result with equality holding when
Problem Let . Prove that the following inequality holds – My Solution Set , and similarly that for any , where, we thus have that – Thus, Therefore we have , Thus it is sufficient to prove that – By Holders’ Inequality , Equality occurs when Hence our proof is completed.