Need help with three proofs.

|x+y|

two possibilities
A:
x+y
B:
-x-y = -(x+y)

if A:
x+y is maximum when both are +
then x+y = |x|+|y|

if B:
-(x+y) has same absolute value as A

etc

google "triangle inequality" for many discussions and proofs

Sure! Let's go through each proof step by step.

Proof #1:
To prove that |x+y| is less than or equal to |x|+|y|, we need to consider two cases: when x+y is positive and when it is negative.

Case 1: x+y is positive.
In this case, |x+y| = x+y and |x|+|y| = x+y. Since both quantities are equal, the inequality holds.

Case 2: x+y is negative.
In this case, |x+y| = -(x+y) and |x|+|y| = x+y. Since -(x+y) is always less than or equal to x+y, the inequality holds.

Therefore, regardless of whether x+y is positive or negative, the inequality |x+y| <= |x|+|y| holds true. Proof #1 is complete.

Proof #2:
To prove that |x|-|y| is less than or equal to |x-y|, we can use the triangle inequality.

The triangle inequality states:
|x+y| <= |x| + |y|

Using this inequality, let's substitute x with (x-y) and y with (-y):
|((x-y)-(-y))| <= |x-y| + |-y|

Simplifying the equation:
|(x-y+y)| <= |x-y| + |y|
|x| <= |x-y| + |y|

We can see that |x| is less than or equal to |x-y| + |y|. Since |x|-|y| = |x|, we have proven the desired inequality. Proof #2 is complete.

Proof #3:
To prove that |x+y+z| is less than or equal to |x|+|y|+|z|, we'll apply the triangle inequality multiple times.

Starting with the left side of the inequality:
|x+y+z|

Now, let's split it into two parts:
|(x+y)+z|

Applying the triangle inequality to the first part:
|(x+y)| + |z|

Now, splitting it further to:
|((x+y)-(-z))|

Applying the triangle inequality again:
|((x+y)-(-z))| <= |x+y| + |-z|

Simplifying:
|(x+y+z)| <= |x+y| + |z|

At this point, we have |(x+y+z)| <= |x+y| + |z|. But we can also apply the triangle inequality to the first part again:
|x+y| <= |x| + |y|

Substituting it back:
|(x+y+z)| <= |x| + |y| + |z|

Therefore, |x+y+z| is less than or equal to |x| + |y| + |z|. Proof #3 is complete.

I hope this helps! Let me know if you have any other questions.