Solution for February 2012
The Problem: 

Does there exist a bounded realvalued function $f(x)$ with $f(1) > 0$ such that for all real numbers $x$ and $y$,
$$\left(f(x+y)\right)^2 \ge \left(f(x)\right)^2 + 2f(xy) +\left(f(y)\right)^2 .$$

Solution: No, no such function exists.
Correct solutions were submitted by
Lamis Alsheikh
(Syria) 
Lou Cairoli (USA) 
Bernard Collignon (France) 
Hubert Desprez (France) 
MeiHui Fang (Austria) 
Frank Feys 
Verena Haider (Austria) 
Benoît Humbert (France) 
Omran Kouba (Syria) 
Marc Lichtenberg (France) 
Matthew Lim (USA) 
Albert Stadler (Switzerland) 
Alessandro Ventullo (Italy) 

We also received three incomplete submissions.
Our correspondents established the nonexistence in two different ways.
Method 1. Let $x_0=1$ and define $x_{n+1}=x_n + \frac1{x_n}$. From the given inequality we have
\begin{eqnarray*}
f^2(x_{n+1}) &=& f^2\left(x_n + \frac1{x_n}\right)\\
&\ge& f^2(x_n) +2f(1) + f^2\left(\frac1{x_n}\right) \ge
f^2(x_n) + 2f(1).
\end{eqnarray*}
Applying this result recursively we deduce that
$$f^2(x_n) \ge f^2(1) + 2nf(1),$$
which grows without bound.
Method 2. Suppose to the contrary that $f$ is bounded; then $f^2$ would be bounded above for all real numbers $x$. Denote by $s$ the supremum of $f^2$ over all nonzero $x$; that is, $s=\sup_{x\ne 0}f^2(x)$ is the smallest number for which $f^2(x) \le s$ for all nonzero real numbers $x$. (Recall that the supremum for bounded realvalued functions exists for any nonempty subset of the domain and is unique.) Since the functional inequality implies that for nonzero $x$,
$$f^2\left(x+\frac1x\right) \ge f^2(x) + 2f(1) +f^2\left(\frac1x\right)\ge f^2(x) +2f(1),$$
it follows that for all $x \ne 0$, $x+\frac1x = \frac{x^2+1}{x} \ne 0$, whence
$$s=\sup_{x\ne 0}f^2(x) \ge \sup_{x\ne 0}f^2\left(x+\frac1x\right) \ge \sup_{x \ne 0}\left((f^2(x) + 2f(1)\right) = s + f(1).$$
That is, $s \ge s + f(1)$, which implies that $f(1) \le 0$, contrary to the given inequality for $f(1)$.
Comments. February's problem was $ \#5$ in the 11th form's final round of the XXXI Russian Mathematical Olympiad (2005). The proposer, N. Agakhanov, perhaps chose "2" to be the coefficient of $f(xy)$ (in the problem's inequality) for aesthetic reasons. As Hubert Desprez pointed out in his soulution, it could as well have been 2012, or any other positive number. (He also provided an easy argument for when the problem is modified so that there is no $f(xy)$ term on the right side of the inequality.) He added, as did Lim, that the number 1 likewise plays no essential role in the problem  the condition $f(1) \ge 0$ could be replaced by $f(c) > 0$ for your favorite constant $c$. Some positivity condition is critical though: Lim observed that the constant function $f(x) = 1$ (for all $x$) is bounded and satisfies the functional inequality (but, of course, fails to satisfy $f(1) > 0$); for a nonconstant example Kouba suggests the bounded function $f(x) = \min(1, x)$.
