Inequality with roots
Let be positive real numbers. Prove that
Solution
We apply the AM – GM inequality, thus:
Hence it suffices to prove that which holds because it is equivalent to .
On permutation
Let be vectors of – dimensional Euclidean space such that . Prove that there exists a permutation of the integers such that
for each .
Solution
We define inductively. Set .Assume is defined for and also
(1)
Note that is true for . We choose in a way that is fulfilled for instead of . Set and . Assume that for all . Then and in view of ones gets which is impossible. Hence , there is such that
(2)
Put . Then using and we have
which verifies for . Thus we define for every . Finally from we get
Homomorphism and inequality
Let be a group and be a homomorphism. Prove that
Solution
In general it holds that ( first isomorphism theorem ) . Taking that for granted we also have
and the inequality is equivelant to which is obviously true.
The exercise can also be found at mathematica.gr .
On linear operators
Let and suppose that , are linear operators from into satisfying
(1)
 Show that for all one has
 Show that there exists such that .
Solution
 Using the assumptions we have

Consider the linear operator acting over all matrices . It may have at most different eigenvalues. Assuming that for every we get that has infinitely many different eigenvalues in view of (i). This is a contradiction.
On the sum of inverse binomial
In this post we are discussing the sum
In Staver was the first to study the sum . He observed that . He was , then , able to extract the recursive relation
(1)
He then proved a great result which is well known in literature
(2)
Later, in Rocket combining the identity
(3)
along with induction he was able to give another proof of . In Surin provided another proof using the well known integral representation of the binomial coefficient,
(4)
Since then the cases and have been studied extensively. However, Mansour generalising the idea of Sury provided a theorem which states the following:
Theorem [Mansour]: Let be non negative integers and be given by
where are two functions defined on . Let , be two sequences and , be their corresponding generating functions. Then,
Proof: The proof is a standard generating type one and is left to the reader.
Using the above theorem along with equation we can generate wonderful stuff. For example:
Example 1: Pick and . Then
which , after a bit of transformations gives the general result
Fabulous, isn’t it? If we set we get equation . Of course there are other applications of the above theorem. We can establish a similar equality for the sum . Another relation that can be established by the above theorem is the following:
(5)
We are not gonna go into a deep analysis but the following equalities also hold:
(6)
and
(7)