Thread: Inequality in R

In trying to solve the following inequality in R:

|x+y| |y+z|+|z+x| $\displaystyle \leq$ |x+y+z| |x| +|y| + |z|,i tried a proof by contradiction.

So suppose the above inequality is not true ,then we have:


|x+y| |y+z|+|z+x| > |x+y+z| |x| +|y| + |z|

....................or 2|x| +2|y| +2|z| > |x+y+z| +|x| +|y| +|z|

.....................or |x| +|y| + |z|> |x+y+z| and here is where i get stuck because i cannot derive a contradiction

Originally Posted by psolaki

In trying to solve the following inequality in R:

|x+y| |y+z|+|z+x| $\displaystyle \leq$ |x+y+z| |x| +|y| + |z|
[snip]

Are you trying to prove this?
If so, it isn't true. For example, try x = y = z = 1.

Originally Posted by awkward

Are you trying to prove this?
If so, it isn't true. For example, try x = y = z = 1.

|x+y| +|y+z|+|z+x|=|1+1|+|1+1|+|1+1|= 2+2+2 =6

|x|+|y|+|z| +|x+y+z| = |1|+|1|+|1|+|1+1+1| = 3+3=6

Hence |x+y|+|y+z|+|z+x|$\displaystyle \leq$ |x|+|y|+|z|+|x+y+z| is true

Originally Posted by psolaki

|x+y| +|y+z|+|z+x|=|1+1|+|1+1|+|1+1|= 2+2+2 =6
|x|+|y|+|z| +|x+y+z| = |1|+|1|+|1|+|1+1+1| = 3+3=6
Hence |x+y|+|y+z|+|z+x|$\displaystyle \leq$ |x|+|y|+|z|+|x+y+z| is true

But that is not the problem you posted!
What is the correct problem?

Classic case of bait & switch ! Ha! Ha!

Originally Posted by Plato

But that is not the problem you posted!
What is the correct problem?

Originally Posted by psolaki

In trying to solve the following inequality in R:



|x+y| +|y+z|+|z+x| $\displaystyle \leq$ |x+y+z| |x| +|y| + |z|,i tried a proof by contradiction.



So suppose the above inequality is not true ,then we have:





|x+y| |y+z|+|z+x| > |x+y+z| |x| +|y| + |z|



....................or 2|x| +2|y| +2|z| > |x+y+z| +|x| +|y| +|z|



.....................or |x| +|y| + |z|> |x+y+z| and here is where i get stuck because i cannot derive a contradiction

Here is my opening post

Originally Posted by psolaki

Here is my opening post

We all can read your OP.

Originally Posted by psolaki

In trying to solve the following inequality in R:
$\displaystyle |x+y| |y+z|+|z+x| \leq |x+y+z| |x| +|y| + |z|$,i tried a proof by contradiction.

Clearly it is $\displaystyle |x+y| |y+z|+|z+x| \leq |x+y+z| |x| +|y| + |z|$
Let $\displaystyle x=y=z=1$ then you have
$\displaystyle 6=|1+1| |1+1|+|1+1| \leq |1+1+1| |1| +|1| + |1|=5$
So it is false.

Why can't you see that?
Or if what you posted is not correct, then why not simply correct it?

It is one of those things that no body can explain. I must have gone at least 10 times over my opening post ,yet a terrible mistake has escape my attention .

I am terribly sorry the correct inequality is:

|x+y|+|y+z|+|z+x|$\displaystyle \leq$ |x|+|y|+|z|+|x+y+z|