If \(n\) is a positive integer, we denote by \(s(n)\) the sum of the divisors of \(n\). For example, the divisors of \(n=6\) are \(1,2,3,6\), so \(s(6)=1+2+3+6=12\). Prove that, for all \(n\geq1\), \[s(1)+s(2)+\cdots+s(n)\leq n^2.\] Denote by \(t(n)\) is instead the sum of the squares of the divisors of \(n\) (e.g., \(t(6)=1^2+2^2+3^2+6^2=50\)), can you find a similar inequality for \(t(n)\)?