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#1 2006-10-17 20:53:22

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

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.

Bernoulli's Inequality

For x > -1, x ≠ 0, and integers n > 1,

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[sub]k[/sub],

Proof

Restate the inequality as follows:

Let {x[sub]i[/sub]}[sub]i = 1[/sub][sup]n[/sup] 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[sub]1[/sub] ≤ p[sub]2[/sub], we have the inequality

Since f(p) is positive, write

so that x[sup]p[sub]2[/sub][/sup] = ∑[sub]i = 1[/sub][sup]n[/sup] x[sub]i[/sub][sup]p[sub]2[/sub][/sup]. This implies that x[sub]i[/sub][sup]p[sub]2[/sub][/sup] ≤ x[sup]p[sub]2[/sub][/sup] and thus x[sub]i[/sub] ≤ x for each i. Since p[sub]2[/sub] ≥ p[sub]1[/sub] > 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[sub]ij[/sub] and transpose A[sup]T[/sup]. Then

Proof

Apply the Gram determinant(Gramian), defined as follows:

In other words, the Gramian is the determinant of the matrix with ij-entry 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[sub]1[/sub],..., x[sub]k[/sub]} are k vectors in E[sup]n[/sup] then G(x[sub]1[/sub],..., x[sub]k[/sub]) is the square of the k-dimensional measure of the k-parallelotope determined by the vectors, or in other words, G(x[sub]1[/sub],..., x[sub]k[/sub]) is the square of the determinant of the k × k minor consisting of the i[sub]1[/sub], i[sub]2[/sub],..., i[sub]k[/sub] rows. Then given n vectors in E[sup]n[/sup], G(x[sub]1[/sub],..., x[sub]n[/sub]) = 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.

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#2 2006-10-18 13:45:34

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

Re: Proofs of selected inequalities from the Formulas thread.

Great!


X'(y-Xβ)=0

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