You are not logged in.

- Topics: Active | Unanswered

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

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.

Offline

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

Great!

**X'(y-Xβ)=0**

Offline