# Proof, strong triangle inequality

1. Dec 18, 2006

### frb

can someone explain this proof please, I added a star to the inequalities I don't see/understand.

if | | is a norm on a field K and if there is a C > 0 so that for all integers n |n.1| is smaller than or equal to C, the norm is non archimedean (ie the strong triangle inequality is true)

proof: if x and y in K

$$$\begin{array}{l} |x + y|^n \le \sum\limits_{k = 0}^n {|\frac{{n!}}{{k!(n - k)!}}} x^k y^{n - k} | \le *(n + 1).C.\max \left( {|x|,|y|} \right)^n \\ |x + y| \le *\mathop {\lim }\limits_{n \to \infty } \left[ {(n + 1)C.\max \left( {|x|,|y|} \right)^n } \right]^{1/n} * = \max \left( {|x|,|y|} \right) \\ \end{array}$$$

I understand everything except the parts I marked with a *

Last edited: Dec 18, 2006
2. Dec 18, 2006

### matt grime

The first * seems to assume that the binomial coefficient is less than C in norm, though I don't claim to be able to explain that right now, and is the only thing to do with non-rchmideanness. The second * is just taking roots in the first line. And the third * is just taking the limit. The last is standard - the n'th root of a constant tends to 1 as n tends to infinity, as does the n'th root of n+1.

3. Dec 18, 2006

### frb

thank you very much

Last edited: Dec 18, 2006