# Inequality involving square roots

#### Ruun

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 $$\displaystyle a^2 \geq b^2 \Rightarrow a \geq b$$

#### gmatt

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 $$\displaystyle a < b \implies a^2 < b^2$$?

#### Ruun

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
Looking at this idea from a calculus perspective: the square root function is everywhere increasing and positive on its domain. Therefore, it preserves order.

• Ruun

#### gmatt

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 $$\displaystyle (a^2 -b^2) = (a+b)(a-b) < (a+b) 0 = 0.$$

• Ruun

#### Plato

MHF Helper
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.

• Ruun and Ackbeet

#### Ruun

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
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.