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\begin{document}

\centerline{\large \bf New Zealand Mathematical Olympiad Camp}
\bigskip
\medskip

\centerline{\large \bf Christchurch, 1998}
\bigskip
\medskip

\centerline{\Large \bf TEAM PROBLEMS NO. 1}
\vspace{2cm}

\begin{enumerate}
\item Let $a,b,c$ be positive real numbers. Prove that
$$
\left(1+{a\over b}\right)\cdot\left(1+{b\over
c}\right)\cdot\left(1+{c\over
a}\right)
\ge 2\left(1+{a+b+c\over \sqrt[3]{abc}}\right).
$$

\item Let 
$$
S=\{(x,y,z) \mid x^2+y^2+z^2=1 \}
$$ 
be the unit sphere in space endowed with the orthogonal coordinate
system. 
Prove
that there is a function 
$f\colon S\rightarrow S$ such that
\[
f(f(x,y,z))=(-x,-y,-z).
\]

\item Find all functions $f\colon \R\rightarrow \R$ such that
$$
f(xf(x)+f(y))=(f(x))^2+y
$$
for all real numbers $x$ and $y$.

\end{enumerate}
\newpage
%============================

\centerline{\large \bf New Zealand Mathematical Olympiad Camp}
\bigskip
\medskip

\centerline{\large \bf Christchurch, 1998}
\bigskip
\medskip

\centerline{\Large \bf TEAM PROBLEMS NO. 1}
\vspace{1cm}
\centerline{\Large \bf Solutions}
\vspace{1cm}

{\bf 1.} Let 
$$
x={a\over \sqrt[3]{abc}}, \quad y={b\over \sqrt[3]{abc}}, \quad 
z={c\over \sqrt[3]{abc}}.
$$
The original inequality becomes
$$
\left(1+{x\over y}\right)\cdot\left(1+{y\over
z}\right)\cdot\left(1+{z\over
x}\right)
\ge 2(1+x+y+z),
$$
or, since $xyz=1$, 
$$
(x+y)(y+z)(z+x)\ge 2+2(x+y+z)
$$
which is equivalent to 
$$
(x+y+z)(xy+yz+zx-2)-xyz\ge 2.
$$
This inequality holds because $xyz=1$ and the AM-GM inequality which
implies
$$
x+y+z\ge 3,\qquad xy+yz+zx\ge 3.
$$
\smallskip

{\bf 2.} Let us adopt the geographic terminology and fix two large
circles
which are
situated in two mutually perpedicular planes passing through the centre
of the
sphere. We shall call them the Greenwich meridian and the equator. The
function $f$
will be defined for points on the Greenwich meridian as a rotation
through
the angle
$90^{\circ}$ degrees in one of the two directions. Similarly $f$ will be
defined on
any other meridian with the exception of the points belonging to the
equator. 
This gives us a function $f$ with the property 
\begin{equation}
f(f(x,y,z))=(-x,-y,-z),
\end{equation}
defined for almost all points $(x,y,z)\in \R^3$. The points which are
left
are the
points of the equator without two diametrically opposite points $M$ and
$N$
belonging to the Greenwich meridian (see the figure).\par
\begin{figure}[h]
\centerline{\picture 2.25in by 1.8in (fig6 scaled 800)}
\caption{}
\end{figure}
Let us choose any $4$ points $A,B,C,D$ of the equator which are
different
from $M$ and $N$, and are vertices of the rectangle with one pair of
sides
parallel
to $MN$ and the other pair of sides perpendicular to $MN$. We admit also
the only
degenerate rectangle with $B=C$ and $A=D$. We set $f(A)=B$, $f(B)=C$,
$f(C)=D$,
$f(D)=A$. Since $A=-C$ and $B=-D$ the property (1) will also hold for
the
points of
the equator.\par
\bigskip

{\bf 3.} Let us denote $b=f(0)$ and substitute $x=0$ into the equation 
$$
f(xf(x)+f(y))=(f(x))^2+y. \eqno(1)
$$
We get 
$$
f(f(y))=b^2+y\qquad \mbox{for all }\ y\in \R.
$$
This last equation shows that $f\colon \R \rightarrow \R$ is a
bijection,
and thus
there exists $c$ such that $f(c)=0$. Substituting $x=c$ in (1) we get
$$
f(f(y))=y\qquad \mbox{for all }\ y\in \R. \eqno(2)
$$
Substituting $y=0$ in (1) we get 
$$
f(xf(x))=(f(x))^2. \eqno(3)
$$
and substituting $x=f(y)$ in the latter
$$
f(yf(y))=y^2. \eqno(4)
$$
Comparing (3) and (4) we get $x^2=(f(x))^2$ for all $x\in \R$, i.e.,
$f(x)=\pm x$.
Suppose that there exist $x_0$ and $y_0$ such that 
$$
f(x_0)=-x_0\ \mbox{ and }\ f(y_0)=y_0 \eqno(5)
$$
for some $x_0\ne 0$, $y_0\ne 0$. Then (1) implies
$$
f\left(-x_0^2+y_0\right)=f\left(x_0f(x_0)+f(y_0)\right)=x_0^2+y_0\ne \pm
(-x_0^2+y_0).
$$
Therefore (5) lead us to a contradiction. Thus we have only two
possibilities:
$f(x)=x$ for all $x\in \R$ or $f(x)=-x$ for all $x\in \R$. Both
functions
satisfy
the equation (1).


\end{document}