\section*{Problem 1 - 121241}

Determine whether among all pairs \((a, b)\) of positive real numbers, there exist such pairs for which

\[
f(a, b)=\frac{a^{4}}{b^{4}}+\frac{b^{4}}{a^{4}}-\frac{a^{2}}{b^{2}}-\frac{b^{2}}{a^{2}}+\frac{a}{b}+\frac{b}{a}
\]

attains a minimum value. If so, then this minimum value should be specified.
}

Due to the symmetry of the function, instead of \(f(a, b)\), we can consider \(g(x)=x^{4}+\frac{1}{x^{4}}-x^{2}-\frac{1}{x^{2}}+x+\frac{1}{x}\) with \(x:=\frac{a}{b}>0\). It holds that

\[
\begin{aligned}
g(x)-x-\frac{1}{x} & =\left(x^{2}+\frac{1}{x^{2}}\right)^{2}-\left(x+\frac{1}{x}\right)^{2} \\
& =\left(x^{2}+\frac{1}{x^{2}}+x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}-x-\frac{1}{x}\right) \\
& =\left(x^{2}+\frac{1}{x^{2}}+x+\frac{1}{x}\right) \frac{(x-1)^{2}\left(x^{2}+x+1\right)}{x^{2}}
\end{aligned}
\]

The factors in the last line are all non-negative. Only \((x-1)^{2}\) becomes zero for \(x=1\). Thus, we obtain \(g(x) \geq x+\frac{1}{x} \geq 2\), with equality only for \(x=1\). Therefore, \(f\) attains its minimum at \(f(a, a)=2\).

\section*{Source anonymous}