Monthly Archives: February 2010
Problem Prove that for positive reals – Proof Clear the denominators and observe the above inequality is equivalent to – which is the schur’s inequality for degree 1 . Hence proved with equality for or Advertisements
Prove that for side-lengths of a triangle then prove that the following inequality holds , Note : The inequality as proposed originally holds for all positive reals….but I amn’t good enough to solve it without the extra condition I have imposed. Proof Lemma 1 For all positive reals , Lemma 2 For all sidelengths of […]
Problem For positive reals such that , Prove that – Proof The hypothesis implies that – Set , Thus, The inequality is now equivalent to – Since the sum of x,y,z is 1 – Now by AM-GM , Equality occurs when
Problem Prove that for non-negative reals , Proof By AM-GM we have , and also , . Adding the two inequalities we get the desired. Equality occurs when all the variables are equal.
For arbitrary reals . Let – , Prove that the following inequality holds good – Proof note that – By Cauchy-Schwarz Inequality, Hence Proved with equality for . Some results can be proved with the definition –
Problem Prove the following inequality – where for Reals My Solution The inequality is cyclic. So assume that are the middle elements of respectively. By Cauchy-Schwarz Inequality, Therefore it is sufficient to prove that – which is true from the assumption . Hence the inequality is proved with equality for and
It is quite noticeable than this word – “WLOG” (=Without Loss Of Generality) is strewn across the net and books …. This post shall convey its real meaning. (1) If a function – is symmetric . If for some permutation – with Since the function is symmetric, Therefore , Proving, with Thus a pairwise order […]