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.

*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:

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:

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