(NIMO 2014) Looks like Cauchy-Schwarz!

I claim that $f(n^2)=f(n)f(n-1)$ for all $n\geq 1$ . To prove this, write $f(n^2)=n^4+n^2+1=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1).$ Now remark that $f(n-1)=(n-1)^2+(n-1)+1=n^2-n+1$ , so we indeed have $f(n^2)=f(n)f(n-1)$ as desired.

As a result, dividing through by $f(1^2)f(2^2)\ldots f(n^2)$ and engaging in rampant cancellation yields $\begin{aligned}2015&\geq\dfrac{f(1)^2\cdot f(2)^2\cdot\ldots\cdot f(n)^2}{f(1^2)f(2^2)\ldots f(n^2)}\\&=\dfrac{f(1)^2f(2)^2\ldots f(n)^2}{[f(0)f(1)][f(1)f(2)][f(2)f(3)]\ldots [f(n-1)f(n)]}\\&=\dfrac{f(1)^2f(2)^2\ldots f(n)^2}{f(0)f(1)^2f(2)^2\ldots f(n-1)^2f(n)}\\&=\dfrac{f(n)}{f(0)}=n^2+n+1.\end{aligned}$ Hence it suffices to find the smallest $n$ such that $2015\geq n^2+n+1$ . To do this, remark that $n^2+n+1\approx n(n+1)$ . From here, one can see that $44\cdot 45=1980$ and that $45\cdot 46$ is too large, so $n=\boxed{44}$ is the largest value of $n$ which makes the inequality true.