Is there a simpler method to prove absolute inequalities?

[imranmeghji]

prove the folowing and state when the inequality holds...

|x+y+z|<=|x|+|y|+|z|

i was thinking that i consider all the possible cases, ie x is positive, y positive, z positive; then the various combinations with negative as well...
is there another shorter method of doing it?
help...

 

[mathman]

You can start with 2 terms first. |x+y|<=|x|+|y|. (I am assuming you are talking about real numbers). Equal if the same sign, less if opposite. Then the 3 term case can be proven in two steps using the 2 term case.

 

[tacman]

Yes, there is a simpler method to prove absolute inequalities. Instead of considering all possible cases, we can use the triangle inequality property of absolute values.

The triangle inequality states that for any real numbers a and b, |a + b| <= |a| + |b|. This means that the absolute value of the sum of two numbers is always less than or equal to the sum of their absolute values.

Using this property, we can prove the given inequality by breaking it down into smaller parts.

First, we can rewrite the left side of the inequality as |x + (y + z)|. Then, using the triangle inequality property, we can say that |x + (y + z)| <= |x| + |y + z|.

Next, we can again apply the triangle inequality to the right side, breaking it down into |x| + |y| + |z|.

Finally, we can combine the two parts to get |x + (y + z)| <= |x| + |y| + |z|.

This shows that the original inequality holds for all real numbers x, y, and z.

In summary, instead of considering all possible cases, we can use the triangle inequality property to prove absolute inequalities in a simpler and more efficient way.