This seems very easy, unless I messed up. Firstly, we have \[f\left(\frac{1}{x}\right)\geq x^2f(x)\]and now replace \(x\) with \(\frac{1}{x}\) to get that \[x^2f(x)\geq f\left(\frac{1}{x}\right)\]so we get that \[f\left(\frac{1}{x}\right)=x^2f(x)\]Therefore, \[1-f(x)=x^2f(x)\]implying that \(f(x)=\frac{1}{x^2+1}\) for all \(x\).