A direct consequence of rearrangement, Chebyshev is a nice inequality that is able to deduce nicer results and can be extremely powerful sometimes too. This can be proved by rearrangement so this holds for all real numbers.
We already know it from our previous post; if and are similarly sorted sequences then
And if these sequences are oppositely sorted then
Since we already know it’s proof; I will directly go to applications of this inequality in various problems. We have already proven Iran 1996 using Chebyshev’s inequality. I have proved an inequality in Mathematical reflections this issue using simple Chebyshev; and I will post the inequality and my solution using Chebyshev after the deadline (1st November 2009) .We also have a little introduction to Some applications because of my previous post.So now I will have a short introduction to this inequality by deducing some known easy inequalities at first.
We will prove for we have in another way.
We have, the sequences and are similarly sorted so from Chebyshev;
(From Cauchy Schwartz inequality)
Because from AM-GM we have .
Hence we are done
Let be positive real numbers .Prove that:
By Chebyshev,we have
Summing up these inequalities,we get the desired result.
Let be positive reals such that these satisfy ,
(Using Chebyshev’s inequality)
(Serbia 2008)Let , , be positive real numbers such that . Prove that
We have the equivalent inequality :
Now we have the comment:
Because of so that
Without loss of generality, we assume
But we also have the two monotonous sequences ( verify )
(a) ( Note that
Applying Chebyshev inequality for two sequences above, we have
By Chebyshev inequality we have now
Let ; and are positive reals numbers. Show that:
and are same ordered.
from Cauchy-Schwartz inequality.
Prove that for arbitrary non-negative real numbers with sum we have
Assume that . It can be esily proved that
Therefore applying Chebyshev’s Inequality we have
(From AM-GM inequality).
Let be distinct positive integers, then prove
Suppose that ,hence for all
Put ,the inequality becomes:
Then by Chebyshev inequality we have:
We are done.
Let and , Prove that:
Let . Then using , our inequality is equivalent to
For the left part, use , which is true since .
Now we come to the right part.
Let so that
Hence we can apply Chebyshev:
Let be a,b and c non-negative reals numbers. Show that
rewrite the given inequality as:
after adding to , etc. Now the above inequality is evident by Chebyshev.
Let be real numbers. Prove that:
Which is obvious from chebyshev’s inequality.
Equality holds iff
WLOG . then from Chebyshev we have:
so we need to show that
which is just the trivial inequality. Note that the right side also follows from the last inequality.
Let be positive real numbers, show that
We claim that
Hint for proof : You can use Iran 1996 Inequality. This remains as a challenge for the readers.
Hence, it suffices to show that
Now, WLOG, we may assume that and thus,
Again, using Chebyshev inequality for these and our lemma above the proof is quite obvious.
Now since the deadline for Mathematical Reflections is over (1st Nov. 2009) hence I will post the inequality and my solutions for it.
See my attachment, it is the same document that I sent to MR, but unfortunately got no reply……
You will also get a fine demonstration of using rearrangement nicely.
Please enjoy. (Also tell me if you like my solutions)
* Link for Mathematical reflections 2009 issue 5 submissions:
(To be updated)
1. A generalization of Chebyshev’s inequality. :http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244168