67th Romanian Mathematical Olympiad

Multiplying out yields $$\begin{array}{rl}f(x)& =(1+a_1^x)\cdot (1+a_2^x)\cdot \cdots \cdot (1+a_n^x)\\ & =1+\sum _{1\le i\le n}{a}_i^x+\sum _{1\le i<j\le n}{a}_i^x{a}_j^x+\cdots +a_1^x{a}_2^x\dots a_n^x.\end{array}$$

Since $a_1\cdot a_2\cdot \cdots \cdot a_n=1$, the function $f$ is a sum of terms of the type $a^x+\frac{1}{a^x}$, where $a=a_{i_1}{a}_{i_2}\dots a_{i_k}$, with $k\le n$. Now, for $a>0$, we consider the function $g:(0,\infty )\to \mathbb{R}$, $g(x)=a^x+a^{-x}$. Let $t>0$; then $$\begin{array}{rl}g(x+t)-g(x)& =\frac{a^{2x+2t}+1}{a^{x+t}}-\frac{a^{2x}+1}{a^x}=\frac{a^{2x+2t}-a^{2x+t}-a^t+1}{a^{x+t}}=\\ & =\frac{(a^{2x+t}-1)(a^t-1)}{a^{x+t}}\ge 0,\end{array}$$ hence $g$ is increasing. The conclusion is obvious.