60th Belarusian Mathematical Olympiad

Note that if $d$ takes on the values of all divisors of positive integer $n$, then $n/d$ takes on the values of all divisors of $n$, too. Let $d_1,\dots ,d_{\tau (n)}$ denote the divisors of $n$. We have $$\sigma (n)=\frac{1}{2}\left(\left(d_1+\frac{n}{d_1}\right)+\left(d_2+\frac{n}{d_2}\right)+\cdots +\left(d_{\tau (n)}+\frac{n}{d_{\tau (n)}}\right)\right).(\ast )$$ We prove that $$\frac{\sigma (n)}{\tau (n)}\le \frac{n+1}{2}.(1)$$ First, we show that $x+\frac{n}{x}<y+\frac{n}{y}$, if $0<y<x\le \sqrt{n}$. Indeed, this inequality is equivalent to the inequality $x-y<\frac{n(x-y)}{xy}$, i.e. $xy<n$. The last inequality is valid because $xy<\sqrt{n}\cdot \sqrt{n}=n$. Since either $d_i\le \sqrt{n}$, or $n/d_i\le \sqrt{n}$, we obtain $d_i+\frac{n}{d_i}\le 1+\frac{n}{1}=n+1$. Therefore, substituting $n+1$ for $d_i+n/d_i$ in $(\ast )$, we get $\sigma (n)\le \frac{1}{2}(n+1)\tau (n)$. Inequality (1) is proved. From the proof it follows that the equality is achieved only if $n$ is a prime (otherwise there exist $d_i\ne 1,n$, and then $d_i+\frac{n}{d_i}<n+1$).

Now we prove that $$\frac{\sigma (n)}{\tau (n)}\ge \sqrt{n}.(2)$$ By Cauchy's inequality, $x+\frac{n}{x}\ge 2\sqrt{n}$ for any $x>0$. Therefore, substituting $2\sqrt{n}$ for $d_i+n/d_i$ in $(\ast )$, we get $\sigma (n)\ge \frac{1}{2}\cdot 2\sqrt{n}\tau (n)=\sqrt{n}\tau (n)$. Inequality (2) is proved. From the proof it follows that the equality is achieved only if $n=1$.