Problem #PRU-5188

Problem

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 $n\ge 1$, $$s(1)+s(2)+\cdots +s(n)\le n^2.$$ 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 $t(n)$?