Inequality in R

• Oct 13th 2011, 11:03 AM
psolaki
Inequality in R
In trying to solve the following inequality in R:

|x+y| |y+z|+|z+x| $\leq$ |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
• Oct 13th 2011, 05:34 PM
awkward
Re: Inequality in R
Quote:

Originally Posted by psolaki
In trying to solve the following inequality in R:

|x+y| |y+z|+|z+x| $\leq$ |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.
• Oct 14th 2011, 05:07 AM
psolaki
Re: Inequality in R
Quote:

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| $\leq$ |x|+|y|+|z|+|x+y+z| is true
• Oct 14th 2011, 05:19 AM
Plato
Re: Inequality in R
Quote:

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

But that is not the problem you posted!
What is the correct problem?
• Oct 14th 2011, 05:53 AM
SammyS
Re: Inequality in R
Classic case of bait & switch ! Ha! Ha!
• Oct 15th 2011, 02:11 AM
psolaki
Re: Inequality in R
Quote:

Originally Posted by Plato
But that is not the problem you posted!
What is the correct problem?

Quote:

Originally Posted by psolaki
In trying to solve the following inequality in R:

|x+y| +|y+z|+|z+x| $\leq$ |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
• Oct 15th 2011, 02:54 AM
Plato
Re: Inequality in R
Quote:

Originally Posted by psolaki
Here is my opening post

Quote:

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

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

Why can't you see that?
Or if what you posted is not correct, then why not simply correct it?
• Oct 15th 2011, 02:23 PM
psolaki
Re: Inequality in R
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| $\leq$ |x|+|y|+|z|+|x+y+z|