Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

• Index
•  » Euler Avenue
•  » Proofs of selected inequalities from the Formulas thread.

#1 2006-10-18 18:53:22

Zhylliolom
Real Member

Offline

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 ak,

Proof

Restate the inequality as follows:

Let {xi}i = 1n 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 < p1 ≤ p2, we have the inequality

Since f(p) is positive, write

so that xp2 = ∑i = 1n xip2. This implies that xip2 ≤ xp2 and thus xi ≤ x for each i. Since p2 ≥ p1 > 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.

Let A be an n × n matrix with entries aij and transpose AT. 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 {x1,..., xk} are k vectors in En then G(x1,..., xk) is the square of the k-dimensional measure of the k-parallelotope determined by the vectors, or in other words, G(x1,..., xk) is the square of the determinant of the k × k minor consisting of the i1, i2,..., ik rows. Then given n vectors in En, G(x1,..., xn) = 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 2006-10-19 11:45:34

George,Y
Super Member

Offline

Re: Proofs of selected inequalities from the Formulas thread.

Great!

X'(y-Xβ)=0

• Index
•  » Euler Avenue
•  » Proofs of selected inequalities from the Formulas thread.