# Mathematical Analysis Proof: |x-y|<= |x|+|y|

## Homework Statement

1. Show that for all real numbers x and y:
a) |x-y| ≤ |x| + |y|

## Homework Equations

Possibly -|x| ≤ x ≤ |x|,
and -|y| ≤ y ≤ |y|?

## The Attempt at a Solution

I tried using a direct proof here, but I keep getting stuck, especially since this is my first time ever coming across mathematical analysis, so I don't have much intuition for it. What I did think was possibly trying to use the triangle inequality somehow?
So far I have added the two above equations together to get
-|x| - |y| ≤ x + y ≤ |x| + |y|
But If that is right then I have no idea how to manipulate it to get the original statement to be proven. Any help would be much appreciated as I seem to keep going in circles.

Thanks :)

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fresh_42
Mentor
Since $|x-y|=|y-x|$ you may assume that $x\geq y$ which reduces the cases you have to consider. And this is the triangle inequality, because $|x-y|=|x+(-y)|\leq |x|+|-y|=|x|+|y|$ so you cannot use what you want to prove.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

1. Show that for all real numbers x and y:
a) |x-y| ≤ |x| + |y|

## Homework Equations

Possibly -|x| ≤ x ≤ |x|,
and -|y| ≤ y ≤ |y|?

## The Attempt at a Solution

I tried using a direct proof here, but I keep getting stuck, especially since this is my first time ever coming across mathematical analysis, so I don't have much intuition for it. What I did think was possibly trying to use the triangle inequality somehow?
So far I have added the two above equations together to get
-|x| - |y| ≤ x + y ≤ |x| + |y|

Thanks :)
As fresh_42 points out, you may as well assume that $x < y$; the case $x = y$ is trivial. Furthermore, the cases $x=0$ or $y = 0$ are trivial as well. Given all that, one way to go is by a (tedious?) consideration of cases:
(a) $0 < x < y$; (b) $x < 0 < y$; and (c) $x < y < 0$.

StoneTemplePython
Gold Member
2019 Award
I think the 'right' way to do this is proving Cauchy Schwarz first, then proving triangle inequality.
- - - -
A less often used approach that I happen to like is quasi-linearisation. (Note, change of variables $z:= -y$)

you want to prove

$f \big( x + z \big) = \big \vert x + z \big \vert \leq \big \vert x \big \vert + \big \vert z \big \vert = f \big( x \big) + f \big( z \big)$

now here I've defined $f$ as the absolute value function -- in common shorthand, I've said $f(u) := \big \vert u\big \vert$

\begin{equation*} \begin{aligned} & f\big(u\big) := {\text{maximize }}\lambda \cdot u\\ & \text{subject to: }\lambda \in \{-1,+1\}\\ \end{aligned} \end{equation*}

From here, you'd need to confirm that the two definitions are equivalent, and the obvious fact that 2 choices are better than 1 (at least in 'single player' optimization problems).

MathematicalPhysicist
Gold Member
The same old question year after year.
:-)

Thank you all for your replies :)

Ray Vickson
Homework Helper
Dearly Missed
I think the 'right' way to do this is proving Cauchy Schwarz first, then proving triangle inequality.
- - - -
A less often used approach that I happen to like is quasi-linearisation. (Note, change of variables $z:= -y$)

you want to prove

$f \big( x + z \big) = \big \vert x + z \big \vert \leq \big \vert x \big \vert + \big \vert z \big \vert = f \big( x \big) + f \big( z \big)$

now here I've defined $f$ as the absolute value function -- in common shorthand, I've said $f(u) := \big \vert u\big \vert$

\begin{equation*} \begin{aligned} & f\big(u\big) := {\text{maximize }}\lambda \cdot u\\ & \text{subject to: }\lambda \in \{-1,+1\}\\ \end{aligned} \end{equation*}

From here, you'd need to confirm that the two definitions are equivalent, and the obvious fact that 2 choices are better than 1 (at least in 'single player' optimization problems).
You could even say
$$f(u) = \max_{-1 \leq \lambda \leq 1} \lambda \cdot u$$
to make it "less combinatorical".

StoneTemplePython
Gold Member
2019 Award
You could even say
$$f(u) = \max_{-1 \leq \lambda \leq 1} \lambda \cdot u$$
to make it "less combinatorical".
It definitely could be done as $\lambda \in [-1,1]$.

- - - -
When using quasi-linearisation, it's common to have the 'norm' associated with the constraint fixed to 1 (e.g. when using the technique to pass from Hoelder $\to$ Minkowski).

I originally wrote the constraint as $\big \vert \lambda \big \vert = 1$, but that struck me as perverse given that I was redefining the absolute value function here!