# Chebyshev Inequality

Original Post.

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.
Statement
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.

Nessbitt’s inequality
We will prove for we have in another way.
We have, the sequences and are similarly sorted so from Chebyshev;

(From Cauchy Schwartz inequality)

2. (Problem by Darij Grinberg) For prove that

Since sequences and are similarly sorted hence from Chebyshev;

(from Iran 1996)

Since
So we are done.

Applications
Let denote the lengths of the three sides opposite to the three angles of a triangle.
Show that:

Solution
Let so . By Chebyshev inequality we have:

For the right inequality, note that:

which is true for sides of a triangle.

For Prove that

Solution

Sequences and are similarly sorted.Hence from Chebyshev we have

Let ; prove that :

Solution
Observe that sequences and are similarly sorted (verify ). Hence we can apply Chebyshev on these to obtain:

Because from AM-GM we have .
Hence we are done
Let be positive real numbers .Prove that:

Solution
By Chebyshev,we have
.
Summing up these inequalities,we get the desired result.
Let be positive reals such that these satisfy ,

Solution
WLOG assume
Because :
Hence:

(Using Chebyshev’s inequality)
. QED
(Serbia 2008)Let , , be positive real numbers such that . Prove that

Solution
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
(b)
Applying Chebyshev inequality for two sequences above, we have

But it is easy to check by homogeneous method the following inequality:

Let , prove that:
Solution

, which follows from Chebyshev’s inequality .

. For , show that

Solution
Because
so,(by Chebyshev’s inequality, why? )
Find the maximum value of following expression
İf are positive real numbers and
Solution(Honey_S)
Rewrite the inequality as:

Assume then we have

Thus we have by Chebyshev inequality

We finish our proof here
Prove that where
Solution
Using Chebyshev’s inequality

.
Prove that for we always have:

Solution
Assume WLOG that . Then we have , and
Now by Chebyshev we have:

. Prove that for positive reals , we have:

Solution
For positive reals we have , so we have:

Assume
so we have and

By Chebyshev inequality we have now

. QED
Let ; and are positive reals numbers. Show that:

Solution
and are same ordered.
Hence,

from Cauchy-Schwartz inequality.
Prove that for arbitrary non-negative real numbers with sum we have

Solution
Assume that . It can be esily proved that

Therefore applying Chebyshev’s Inequality we have

and

The rest is trivial. Hence we are done
Let ; Prove that :

Solution
From Chebyshev, since sequences and are similarly sorted hence we have :

(From AM-GM inequality).
Done.
Let be distinct positive integers, then prove

Solution
Suppose that ,hence for all
Put ,the inequality becomes:

Note that
Then by Chebyshev inequality we have:

We are done.
Let and , Prove that:

Solution
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:

QED.
Let be a,b and c non-negative reals numbers. Show that

Solution
rewrite the given inequality as:

after adding to , etc. Now the above inequality is evident by Chebyshev.

Let be real numbers. Prove that:

Solution
Rewrite the inequality as:

But we have; . So we are only required to prove that :

Which is obvious from chebyshev’s inequality.
Equality holds iff
If ,
Solution
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

Solution
Lemma
We claim that

Hint for proof : You can use Iran 1996 Inequality. This remains as a challenge for the readers.
We have

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:
http://reflections.awesomemath.org/currentissue.html
_______________________
(To be updated)
References
1. A generalization of Chebyshev’s inequality. :http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244168