Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

  • 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) > 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.

Hadamard's Inequality

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) = D. 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.

Board footer

Powered by FluxBB