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

• Bonnie
In summary: 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!

## 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 :)

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.

Bonnie said:

## 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##.

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).

The same old question year after year.
:-)

Thank you all for your replies :)

StoneTemplePython said:
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".

Ray Vickson said:
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!

## 1. How do you interpret the statement |x-y|<= |x|+|y| in Mathematical Analysis Proof?

The statement |x-y|<= |x|+|y| means that the absolute value of the difference between two real numbers, x and y, is less than or equal to the sum of the absolute values of x and y.

## 2. Why is it important to prove the statement |x-y|<= |x|+|y| in Mathematical Analysis?

Proving this statement is important because it is a fundamental property in mathematical analysis that allows us to manipulate and solve equations involving absolute values. It also has many applications in various fields such as physics, engineering, and economics.

## 3. What is the general strategy for proving |x-y|<= |x|+|y| in Mathematical Analysis?

The general strategy for proving this statement involves breaking down the absolute value expressions into different cases based on the signs of x and y. Then, using basic properties of real numbers, such as the triangle inequality, to manipulate and simplify the expressions until the statement is proven for all possible cases.

## 4. Are there any special cases or exceptions to the statement |x-y|<= |x|+|y|?

Yes, there are a few special cases where the statement may not hold. For example, when x and y are both equal to 0, the statement becomes 0 <= 0+0, which is true. However, when x and y are both equal to a negative number, the statement becomes |x-y|<= |x|+|y| = -x-y <= -x-y, which is not always true.

## 5. Can you provide an example of a proof for |x-y|<= |x|+|y| in Mathematical Analysis?

Yes, here is a simple example of a proof for this statement:

Proof: Let x and y be any real numbers.
Case 1: If x>=0 and y>=0, then |x-y|=x-y and |x|+|y|=x+y. Therefore, x-y <= x+y, which is true.
Case 2: If x<=0 and y<=0, then |x-y|=-(x-y) and |x|+|y|=-x-y. Therefore, -(x-y) <= -x-y, which is equivalent to x-y <= x+y, and is also true.
Case 3: If x>=0 and y<=0, then |x-y|=x+y and |x|+|y|=x-y. Therefore, x+y <= x-y, which is not always true.
Case 4: If x<=0 and y>=0, then |x-y|=-(x+y) and |x|+|y|=-x+y. Therefore, -(x+y) <= -x+y, which is equivalent to x+y <= x-y, and is also true.
Hence, we have proven the statement for all possible cases and it is true for all real numbers x and y. Therefore, |x-y|<= |x|+|y| is true in general.