Prove that for all $n\in\mathbb{N}$, $\frac{s(n)}{d(n)}\geq \sqrt n$ 
Prove that for all $n\in\mathbb{N}$ $$\frac{s(n)}{d(n)}\geq \sqrt n$$ where $s(n) = \sum_{d|n} d$ and $d(n) = \sum_{d|n} 1$.

Being honest, study some time arithmetic functions, and can not enchegar a means of solving this demonstration ... I searched the internet but nothing has been found (in Portuguese) so I could solve a this.
 A: Hint: Take the natural pairing of the divisor and show that $ \frac{d + \frac{n}{d} } {2} \geq \sqrt{n}$.
Deal with the case where $n$ is a perfect square separately, for completeness.
A: According to the fundamental theorem of arithmetic, any natural number N can be written as a unique product of powers of prime numbers, or $N=\displaystyle\prod_{j=1}^kp_j^{n_j}$ . It can then be shown that

See here and here for proofs of these two relations. Then use them together with the inequality of arithmetic and geometric means, and write $\displaystyle{\frac{D_S}{d(N)}\geqslant\sqrt[d(N)]{D_P}=\sqrt[d(N)]{N^\frac{d(N)}2}=N^\frac12=\sqrt N}$ . QED.
A: Hint: $\frac{s(n)}{d(n)}$ is the arithmetical mean of the divisors of $n$ (Which means you can use AM-GM). However, i managed to improve the bound to $\frac{s(n)}{d(n)}\geq\frac{3}{2}\sqrt{n}-1$ by using $d(n)\leq 2\sqrt n$ and proving: $s(n)\geq\sqrt n.d(n)+(\sqrt n-1)^2$
If you are interested about these bounds, for composite $n>4$ one can prove that $\frac{3}{8}n+1\geq\frac{s(n)}{d(n)}\geq\frac{3}{2}\sqrt{n}-1$
