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You are not logged in. #1 20061018 18:53:22
Proofs of selected inequalities from the Formulas thread.Upon request, I have supplied proofs to some of the inequalities in the Inequalities thread in the Formulas section. I have selected those inequalities which seem less known, so feel free to request some other proofs. Proof Use induction. The base case is n = 2, which gives (1 + x)² > 1 + 2x, which is certainly true upon expanding the left hand side. Now assume the inequality holds for some n ≥ 2. Then since x > 1, 1 + x > 0 and we can write which again is certainly true upon expanding the left hand side. Thus the proof is complete. Jensen's Inequality For 0 < p ≤ q and positive a_{k}, Proof Restate the inequality as follows: Let {x_{i}}_{i = 1}^{n} be a fixed set of n positive numbers and p > 0. The function f(p) defined on (0, ∞) by is positive decreasing function, and therefore if 0 < p_{1} ≤ p_{2}, we have the inequality Since f(p) is positive, write so that x^{p2} = ∑_{i = 1}^{n} x_{i}^{p2}. This implies that x_{i}^{p2} ≤ x^{p2} and thus x_{i} ≤ x for each i. Since p_{2} ≥ p_{1} > 0, it follows that From the above analysis, the left hand side of this inequality is simply 1, so we now have which immediately gives which is the desired result. Hadamard's Inequality Let A be an n × n matrix with entries a_{ij} and transpose A^{T}. Then Proof Apply the Gram determinant(Gramian), defined as follows: In other words, the Gramian is the determinant of the matrix with ijentry equal to the inner product of the ith and jth vector argument of G. Now a theorem(whose proof is excluded at least in this discussion) on Gramians states that if {x_{1},..., x_{k}} are k vectors in E^{n} then G(x_{1},..., x_{k}) is the square of the kdimensional measure of the kparallelotope determined by the vectors, or in other words, G(x_{1},..., x_{k}) is the square of the determinant of the k × k minor consisting of the i_{1}, i_{2},..., i_{k} rows. Then given n vectors in E^{n}, G(x_{1},..., x_{n}) = D². Then it follows that Since the determinant of the transpose is also D, the indices i and j may be interchanged and the inequality is the same. There are a few proofs for now, it is time for me to go. Perhaps later more will be added. #2 20061019 11:45:34
Re: Proofs of selected inequalities from the Formulas thread.Great! X'(yXβ)=0 