Ytukyg

$$a^2+b^2+2|a||b|geq a^2+b^2+2ab$$
$$(|a|+|b|)^2 geq |a+b|^2phantom{a}(because forall xin mathbb{R};phantom{;}x^2=|x|^2)$$
$$therefore |a|+|b|geq |a+b|$$

If a neat algebraic argument does not suggest itself, we can do a crude argument by cases, guided by the examples $a=7,b=4$, $a=-7,b=-4$, $a=7, b=-4$, and $a=-7, b=4$.

If $age 0$ and $bge 0$ then $|a+b|=|a|+|b|$.

If $ale 0$, and $ble 0$, then $|a+b|=-(a+b)=(-a)+(-b)=|a|+|b|$.

Now we need to examine the cases where $a$ is positive and $b$ is negative, or the other way around. Without loss of generality we may assume that $|b|le |a|$.

If $agt 0$, then $|a+b|=|a|-|b|$. This is $lt |a|$, and in particular $lt |a|+|b|$.

If $alt 0$, then again $|a+b|=|a|-|b|$.

A simple proof of the triangle inequality that is complete and easy to understand (there are more cases than strictly necessary; however, my goal is clarity, not conciseness).

Prove the triangle inequality $| x | + | y| ≥ | x + y|$.

Without loss of generality, we need only consider the following cases:

1. $x = 0$


2. $x > 0, y > 0$


3. $x > 0, y < 0$

Case $1$. Suppose $x = 0$. Then we have

$| x| = 0$

$| x| + | y| = 0 + | y| = | y|$

Thus $| x| + | y| = | x + y|$.

Case $2$. Suppose $x > 0, y > 0$. Then, since $x + y > 0$, we have

$| x| = x$

$| y| = y$

$| x| + | y| = x + y$

$| x + y| = x + y$

Thus $| x| + | y| = | x + y|$.

Case $3$. Suppose $x < 0, y < 0$. Then, since $x + y < 0$, we have

$| x| = −x$

$| y| = −y$

$| x| + | y| = (−x) + (−y)$

$| x + y| = −(x + y) = (−x) + (−y)$

Thus $| x| + | y| = | x + y|$.

Case $4$. Suppose $x > 0, y < 0$. Then we have

$| x| = x$

$| y| = −y$

$| x| + | y| = x + (−y)$

We must now consider three cases:

a. $x + y = 0$

b. $x + y > 0$

c. $x + y < 0$

Case $4a$. Suppose $x + y = 0$. Then we have

$| x + y | = |0| = 0$

Since $y < 0$, it follows that $−y > 0$ and thus $x + (−y) > 0 + (-y) = -y > 0$.

Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.

Case $4b$. Suppose $x + y > 0$. Then we have

$| x + y| = x + y$

Since $y < 0$, it follows that $−y > 0 > y$ and thus $x + (−y) > x + y$.

Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.

Case $4c$. Suppose $x + y < 0$. Then we have

$| x + y| = −(x + y) = (−x) + (−y)$

Since $x > 0$, it follows that $x > 0 > −x$ and thus $x + (−y) > (−x) + (−y)$.

Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.

This concludes the proof.

The proof given in Wikipedia / Absolute Value is interesting and the technique can be used for complex numbers:

Choose $epsilon$ from ${ -1,1}$ so that $epsilon (a+b) ge 0$. Clearly $epsilon x le |x|$ for all real $x$ regardless of the value of $epsilon$, so

$|a+b|= epsilon (a+b) = epsilon a + epsilon b le |a|+|b|$

Firstly, observe that $-|x|leq xleq|x|$
, and $-|y|leq yleq|y|$ follow from the definition of the absolute value function. (Consider the cases $x$ is non-negative and $x$ is negative and what happens to $|x|$, the same goes for $y$ mutatis mutandis).

Since $yleq|y|$ , then

$$|x|+yleq|x|+|y|tag{1}$$

by adding $|x|$ to both sides of $yleq|y|$ . Likewise, adding $y$ to both sides of $x leq |x|$ we have:
$$y+x le y+|x|tag{2}$$

Combining equations $1$ and $2$, and using the transitive property of the relation $leq$, we have:

$$y+x le y+|x|leq |y|+|x|$$
$$y+xleq |y|+|x|$$

Likewise, since $-|y|leq y$
, then $-|y|-|x|leq y-|x|$ by adding $-|x|$ to both sides. But $-|x|leq ximplies y-|x|leq y+x$ by adding $y$ to both sides. Thus by the transitive property:

$$ -|y|-|x|leq y-|x|,text{ and }y-|x|leq y+x implies -|y|-|x|leq y+x tag{3}$$

By combining equations 2 and 3, we have:

$$-(|y|+|x|)leq y+xleq|y|+|x|$$

Now noting that $|b|leq aiff-aleq bleq a$
, we have:

$$|x+y|leq|x|+|y|$$

I hope that the following proof is the shortest one and make use of the order in $mathbb{R}$.

• If $abgeq 0$ then
  $$
  |a+b|=
  begin{cases}
  a-b & ifquad ageq 0,; bgeq 0,\
  -a-b & ifquad aleq 0,; bleq 0
  end{cases}
  =|a|+|b|
  $$




• If $ab<0$ then
  $$
  |a+b|leqmax{|a|,|b|}leq |a|+|b|.
  $$

|x+y|^2=(x+y).(x+y)
=(x.x)+2(x.Y)+(y.y)

=|x|^2+2(x.Y)+|y|^2
from Cauchy-Schwarz inequality,|x.Y|<=|x||y|
|x+y|^2 <=|x|^2+2|x||y|+|y|^2
|x+y|^2 <=(|x|+|y|)^2
taking square root on both sides.
|x+y|<=(|x|+|y|)

This proof works alongside the geometric notion that adding numbers on the real line is a 'vector operation'. It also lays out the exact conditions under which the triangle inequality is an equation,

$quad |x + y| = |x| + |y|$.

Recall that in general,

$tag 1 a le b ; text{ iff } ; exists , u ge 0 text{ such that } a + u = b$

It is easy to see that whenever $x, y ge 0$ or $x, y le 0$ the triangle inequality holds since there is
no 'less than' there - $|x+y| = |x| + |y|$.

Logically, there are two case left to be dispensed with:

$quad text{Case 1: } left[x lt 0 lt yright] text{ and } (-x) le y$

$quad text{Case 2: } left[x lt 0 lt yright] text{ and } (-x) ge y$

Since always $|z| = |-z|$, it is only necessary to take care of the first case to prove the triangle inequality.

Case 1

Using $text{(1)}$, we write $(-x) + u = y$ for $u ge 0$ and $(-x) gt 0$. Noting that $u lt y$, we have

$quad |x + y| = |x + (-x) + u| = |u| = u lt y lt (-x) + y = |x| + |y|$

So only when $x$ and $y$ 'straddle $0$' is the triangle inequality a 'strict less than' relation,

$quad |x + y| lt |x| + |y|$.