Another inequality involving sigma and tau

cf. an inequality involving sigma and tau.

Let n be a positive integer. By the weighted AM-GM inequality, one has

\displaystyle\frac{\sum d\cdot n/d}{\sum n/d}\ge\left(\prod d^{n/d}\right)^\frac{1}{\sum n/d},

where all sums and products are taken over the positive divisors d of n. This means

\displaystyle\boxed{\frac{n\tau(n)}{\sigma(n)}\ge\left(\prod_{d\mid n}d^{1/d}\right)^{n/\sigma(n)}}.

Considering the analytic behaviour of the function f(x)=x^{1/x} one deduces

\displaystyle\prod_{d\mid n}d^{1/d}\ge\prod_{1\neq d\mid n}n^{1/n}=n^{(\tau(n)-1)/n}

for any n, with equality iff n is prime or 1 or 4. (By convention we take an empty product to be 1.) Combining this with the first inequality in this post we obtain

\displaystyle n^{1/2}\ge\frac{n\tau(n)}{\sigma(n)}\ge n^\frac{\tau(n)-1}{\sigma(n)},


\boxed{\displaystyle n^{1/2}\le\frac{\sigma(n)}{\tau(n)}\le n^{1-\frac{\tau(n)-1}{\sigma(n)}}}.

This strengthens the first boxed inequality from this post.


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