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I don't understand one step in this proof of the triangle inequality (number 1 below).

I.e. we have the equality $|x+y|^2 =|x|^2+2xy+|y|^2$, but how can we then conclude $|x+y|^2 \leq|x|^2+2|xy|+|y|^2$?

Let $x,y \in \mathbb{R}$ and let $|x|$ be the absolute value of $x$. Then: $$ |x+y|\leq|x|+|y| $$ Proof: \begin{align} |x+y|^2 &= (x+y)^2 \\ &=x^2+2xy+y^2 \\ &= |x|^2+2xy +|y|^2 \\ &\leq |x|^2+2|xy|+|y|^2 \tag{1}\\ &=|x|^2+2|x||y|+|y|^2 \\ &= (|x|+|y|)^2 \end{align}

Thanks!

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  • $\begingroup$ This comes from the fact that absolute value of a real number is greater or equal that number: $\forall t\in \Bbb R \quad x\le |x|$. This can be easily shown by the definition of absolute value. $\endgroup$ – TZakrevskiy Feb 17 '17 at 11:53
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It's because for every number $a\in\mathbb R$, you have $a\leq |a|.$ In particular, here they set $a=xy$.


Longer explanation:

First of all, if $a\in\mathbb R$, then you have two options:

  1. If $a\geq 0$, then $a=|a|$, which means that $a\leq |a|$
  2. If $a<0$, then $|a|>0$ which means that $a<0<|a|$, so $a<|a|$, which again means that $a\leq |a|$.

in total, this means that $a\leq |a|$ is true for any real number $a$.

In particular, this measn that $xy\leq |xy|$ for each value of $x,y\in\mathbb R$.

We also know that if $a\leq b$, then $ac \leq bc$ if $c>0$. In particular, this means that $2xy\leq 2|xy|$.

Now, there is also a rule that if $a\leq b$, then $b+a\leq c+a$ for all $a,b,c\in\mathbb R$. In particular, this means that, because $2xy\leq 2|xy|$, we also know that $$2xy + (|x|^2+|y|^2) \leq 2|xy| + (|x|^2+|y|^2),$$ which is exactly what you need.

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  • $\begingroup$ In point 1, should it not just be "If $a\ge0$, then $a=\left|a\right|$"? $\endgroup$ – lioness99a Feb 17 '17 at 14:04
  • $\begingroup$ @lioness99a Yes, of course. Thank you. $\endgroup$ – 5xum Feb 17 '17 at 14:08
  • $\begingroup$ You also don't need the "which means that..." part of point 1 - we can't conclude that $a\leq\left|a\right|$ from the rest of the statement $\endgroup$ – lioness99a Feb 17 '17 at 14:10
  • $\begingroup$ @lioness99a Of course we can. If $x=y$, then we also know that $x\leq y$. $\endgroup$ – 5xum Feb 17 '17 at 14:11
  • $\begingroup$ I suppose, but in my head that's an odd step. Point 1 proves that $a=\left|a\right|$ and then point 2 further proves that $a<\left|a\right|$. We then combine the two to prove that $a\leq\left|a\right|$. But technically, yes, what you have written is correct $\endgroup$ – lioness99a Feb 17 '17 at 14:13
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Because we can write $2xy\leq|2xy|=2|xy|$.

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