Inequality involving square roots

Mar 2009
129
34
North of Spain
Hi, I've found a proof from \(\displaystyle |x| \leq \sum_{i=1}^{n} |x_{i}|\) given \(\displaystyle x=(x_{1},...,x_{n}), n \in \mathbb{N}\) It reads:

\(\displaystyle \sum_{i=1}^{n}(x_{i})^{2} \leq \sum_{i=1}^{n} (x_{i})^2 + 2\sum_{i \ne j}^{n} |x_{i}||x_{j}| = (\sum_{i=1}^{n}|x_{i}|)^2\).

And taking the square root it's supposed to grant the result.

Now, I do not see why \(\displaystyle a^2 \geq b^2 \Rightarrow a \geq b\)

Thanks in advance
 
Oct 2009
95
29
The statement \(\displaystyle a^2 \ge b^2 \implies a \ge b \) is equivalent to \(\displaystyle a < b \implies a^2 < b^2\) since they are contrapositives. Is it clear to you that \(\displaystyle a < b \implies a^2 < b^2\)?
 
Mar 2009
129
34
North of Spain
Given \(\displaystyle a \leq b \Rightarrow a - b \leq 0\) Now I need information for \(\displaystyle a+b\) because \(\displaystyle (a+b)(a-b) = a^2-b^2\), and that's the piece that I need.

I'm sorry but I still don't get it. Thanks for your post.
 

Ackbeet

MHF Hall of Honor
Jun 2010
6,318
2,433
CT, USA
Looking at this idea from a calculus perspective: the square root function is everywhere increasing and positive on its domain. Therefore, it preserves order.
 
  • Like
Reactions: Ruun
Oct 2009
95
29
Well what you have written is fine. Given \(\displaystyle a < b\) we have that \(\displaystyle a-b < 0\) now provided \(\displaystyle a + b \ge 0\) we have that \(\displaystyle (a^2 -b^2) = (a+b)(a-b) < (a+b) 0 = 0.\)
 
  • Like
Reactions: Ruun

Plato

MHF Helper
Aug 2006
22,508
8,664
Now, I do not see why \(\displaystyle a^2 \geq b^2 \Rightarrow a \geq b\)
I do not understand what exactly you are asking for.
However, \(\displaystyle a^2 \geq b^2 \Rightarrow a \geq b\) is clearly false.
It is true that \(\displaystyle a^2 \geq b^2 \Rightarrow |a| \geq |b|\).

Is it also the case that \(\displaystyle \left| x \right| = \sqrt {\sum\limits_{k = 1}^n {\left( {x_k } \right)^2 } } ~?\)
If so, the proof seems to work.
 
  • Like
Reactions: Ruun and Ackbeet
Mar 2009
129
34
North of Spain
Yes, it is the case.

Well the proof says taking square root, and I didn't find an argument to believe in that, and this is the point I'm looking for,
 

Ackbeet

MHF Hall of Honor
Jun 2010
6,318
2,433
CT, USA
Reply to Plato @ Post #6:

In this case, the things we're taking the square roots of are all positive. The step in question is this one:

\(\displaystyle \sum_{i=1}^{n}(x_{i})^{2} \leq (\sum_{i=1}^{n}|x_{i}|)^2\)

implies

\(\displaystyle \sqrt{\sum_{i=1}^{n}(x_{i})^{2}} \leq \sum_{i=1}^{n}|x_{i}|.\)

It is also true that \(\displaystyle |x|=\sqrt{\sum_{i=1}^{n}(x_{i})^{2}}\), at least in the Euclidean or \(\displaystyle L^{2}\) norm.