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

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!
  • #1
18
1

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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  • #2
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.
 
  • #3
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##.
 
  • #4
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##

But suppose I instead define

##\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).
 
  • #5
The same old question year after year.
:-)
 
  • #6
Thank you all for your replies :)
 
  • #7
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##

But suppose I instead define

##\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".
 
  • #8
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.

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