Yes! It's a very old idea.

Generalization 1 a)Let $n\in \mathbb{N},n\ge 2$, even number .Prove that for any real numbers $a,b,c$ satisfying $abc = 1$ the following inequality holds \[\frac{1}{2{{a}^{n}}+{{b}^{n}}+3n-3}+\frac{1}{2{{b}^{n}}+{{c}^{n}}+3n-3}+\frac{1}{2{{c}^{n}}+{{a}^{n}}+3n-3}\le \frac{1}{n}.\] b)Let $n\in \mathbb{N},n\ge 2$, odd number .Prove that for any positive real numbers $a,b,c$ satisfying $abc = 1$ the following inequality holds \[\frac{1}{2{{a}^{n}}+{{b}^{n}}+3n-3}+\frac{1}{2{{b}^{n}}+{{c}^{n}}+3n-3}+\frac{1}{2{{c}^{n}}+{{a}^{n}}+3n-3}\le \frac{1}{n}.\]

Generalization 2 a)Let $n\in \mathbb{N},n\ge 2$, even number and $k\in {{\mathbb{N}}^{*}}$ .Prove that for any real numbers $a,b,c$ satisfying $abc = 1$ the following inequality holds \[\frac{1}{2k{{a}^{n}}+k{{b}^{n}}+3n-3k}+\frac{1}{2k{{b}^{n}}+k{{c}^{n}}+3n-3k}+\frac{1}{2k{{c}^{n}}+k{{a}^{n}}+3n-3k}\le \frac{1}{n}.\] b)Let $n\in \mathbb{N},n\ge 2$, odd number and $k\in {{\mathbb{N}}^{*}}$.Prove that for any positive real numbers $a,b,c$ satisfying $abc = 1$ the following inequality holds \[\frac{1}{2k{{a}^{n}}+k{{b}^{n}}+3n-3k}+\frac{1}{2k{{b}^{n}}+k{{c}^{n}}+3n-3k}+\frac{1}{2k{{c}^{n}}+k{{a}^{n}}+3n-3k}\le \frac{1}{n}.\]

We will work with $a$ instead of $a^2$ and so on. Now the numbers will be positive. Let $a:=\frac{x}{y}, b:=\frac{y}{z}, c:=\frac{z}{x}$. The inequality becomes \[\sum_{cyc}{\frac{yz}{2xz+y^2+3yz}}\leq \frac{1}{2}\] which can be written as \[\sum_{cyc}(1/3-\frac{yz}{2xz+y^2+3yz})\geq 1/2\] or \[\sum_{cycy}\frac{2xz+y^2}{2xz+y^2+3yz}\geq \frac{3}{2}\] We will use $T2$ and $a^2+b^2+c^2\geq ab+bc+ca$ in what follows. We have \[\sum_{cyc}\frac{y^2}{2xz+y^2+3yz}\geq \frac{\sum{x^2}+2\sum{xy}}{\sum{x^2}+5\sum{xy}}\geq \frac{1}{2}\] and \[\sum_{cyc}\frac{2xz}{2xz+y^2+3yz}\] \[=\sum_{cyc}\frac{(2xz)^2}{(2xz)^2+2xy^2z+6xyz^2}\] \[\geq \frac{\sum{(2xy)^2}+8\sum{xy^2z}}{\sum{(2xy)^2}+8\sum{xy^2z}}=1\] Summing the two inequalities we get the conclusion.

This problem is an old inequality. You can also see problem 11 from the Inequalities marathon; problem proposed by Mircea Lascu. It was also proposed in 2006 for the Romanian National Olympiad.

Indeed, the (version of the) problem, proposed for the 2006 Romanian NMO, is the one mentioned by me in the hidden Solution; identical with that Problem 11 from the Inequalities Marathon that you unearthed at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1626274#p1626274. But what do you mean by "it is a little too easy"? Have you taken a look at Problems 1,2,5 Seniors, and 1,2,5,6 Juniors ? Especially common Problem 5/6 Seniors/Juniors, almost insulting in its lack of any ideas (and with such reduced casework). There is nothing wrong in principle with an easy problem, as long as it's novel, fresh, and not patronizingly obvious.

Sory.I edited that.

I wanted to say that is known and in olympiads known and easy are almost the same thing.

And problems 1,2 and 5 are very easy and boring

Well, "known" and "easy" are indeed the same thing, provided the person taking the test belongs to the set of people for who the problem is "known" If you don't know it, it might not be easy, as it is for the person next to you, who knows it. In other words, knowledge is not universal. I am the last person to embrace usage of known problems ... sometimes old problems are re-discovered. What is wrong is a premeditated use of a known problem (not to mention moral issues of plagiarism and copyright).

You know who was the author of this problem.Was him/her from Romania?I ask this because i found it in at least three or four books in Romania.

All I know is that the name showing in the official booklets of the 2012 Tuymaada as author of the problem is V. Aksenov. Who is the first author, of the first problem of this type, will probably never be known.

mavropnevma wrote: Prove that for any real numbers $a,b,c$ satisfying $abc = 1$ the following inequality holds \[\dfrac{1} {2a^2+b^2+3}+\dfrac {1} {2b^2+c^2+3}+\dfrac{1} {2c^2+a^2+3}\leq \dfrac {1} {2}.\] Proposed by V. Aksenov Let $x$, $y$, $z$ be positive real numbers such that $xyz=1$.Prove that $$\frac{1}{(x+1)^2+y^2+1}+\frac{1}{(y+1)^2+z^2+1}+\frac{1}{(z+1)^2+x^2+1}\le\frac{1}{2}$$