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 […]

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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 🙂

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iSolve : For A,B,C reals Method : Let Choose , rewrite this as – which again is re-written as – Where , Therefore, and are the roots of the equation – Therefore , Now we have that – Solving the quadratic plus the trivial solution gives us – now substituting the value of y in […]

We define is is the exponent of p in the expansion of a. Thus by definition we directly have the following facts – 1. 2. 3. Equality when 4. 5. We also have the legendre’s formula – 6. This function gets to be seriously useful many more plus it is beautiful !! . Setting such […]

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 –