Note that $f$ is differentiable, using product rule it is easy to get that $$f'(x)=\sum_{i=1}^{n} a_i^x \ln a_i\prod_{j\neq i} (1+a_j^x) \Leftrightarrow \frac{f'(x)}{f(x)}=\sum_{i=1}^{n} \frac{a_i^x \ln a_i}{1+a_i^x}$$ Claim. $\frac{a^x \ln a}{1+a^x} \geq \frac{\ln a}{2}$ for all $a,x>0$. Proof. For $a \geq 1$ it is equivalent to $a^x\geq 1$ wich is true, and for $ 0<a<1$ it is equivalent to $a^x\leq 1$ wich is true since $x>0$. By our claim, we have that $$\frac{f'(x)}{f(x)} \geq \sum_{i=1}^{n} \frac{\ln a_i}{2}=\frac{\ln(a_1a_2\dots a_n)}{2}=\frac{\ln 1}{2}=0$$ The last result is the equivalent of $f$ being non-decreasing since $f(x)>0, \forall x>0$. $\blacksquare$