The Lagarias inequality, which is equivalent to the Riemann hypothesis, is:


$$\sigma(n) \le H\_n + \exp(H\_n) \log(H\_n) =:L(n)$$


for all natural numbers $n$, where $\sigma=$ sum of divisors, $H\_n=n$-th harmonic number.
Lagarias inequality is equivalent to:


$$\sigma\left(\frac{ab}{\gcd(a,b)^2}\right) \le L\left(\frac{ab}{\gcd(a,b)^2}\right)$$


for all natural numbers $a,b$. (We see this by plugging in $n=\frac{ab}{\gcd(a,b)^2}$ in one direction, and by setting $a=n,b=1$ in the other direction.)


We write:


$$k(a,b) = \frac{1}{L\left(\dfrac{ab}{\gcd(a,b)^2}\right)}$$


and


$$K(a,b) = \frac{1}{\sigma\left(\dfrac{ab}{\gcd(a,b)^2}\right)}$$


Then the Lagarias inequality is equivalent to: For all natural numbers $a,b$ we have:


$$k(a,b) \le K(a,b)$$


My question is, if $k(a,b),K(a,b)$ are positive definite (obviously symmetric) functions over the natural numbers? (Positive definite kernel: <https://en.wikipedia.org/wiki/Positive-definite_kernel>)


This would allow to interpret the Lagarias inequality as "kernel inequality" in the same way the abc-conjecture is conjectured to be a kernel inequality (see: [The abc-conjecture as an inequality for inner-products?](https://mathoverflow.net/questions/352054/the-abc-conjecture-as-an-inequality-for-inner-products))


[Sagemath computations to back up the conjecture above](https://sagecell.sagemath.org/?z=eJx1j8GKwjAQhu99ihwnNWzJghehJ2UR9KLXskhqYxiymZY0Lj7-ZtMc1GpOw8z3z3zp9IXtQImWrwoWn9fh6onJakTjFKiyrcy5S0BZfvKi6GJguwV65Merg0ZWyC69Z8iQmFdkNEhBC8m_c26_f86tN1-Qti30bZgqXv70Jpc5Z18JxmVzu4Q7BySsra041ofDY86p4PEGTWOnpUlYzYRTu521hUcy9TEe-p_T_Xwp86EhMgEYiewRJeIv-QeOp6EfMeCvPkVLJAwauEjQbjJ9w_A_NFJ6ew==&lang=sage&interacts=eJyLjgUAARUAuQ==)


Thanks for your help!


**Edit**:
It seems that it is better to define:
$$k(a,b) = \frac{1}{L\left(\dfrac{ab}{\gcd(a,b)}\right)}$$


and


$$K(a,b) = \frac{1}{\sigma\left(\dfrac{ab}{\gcd(a,b)}\right)}$$


Then if we let:


$$\phi(n) = \frac{1}{\sigma(n)} \sum\_{d|n} \sqrt{d} e\_d$$
where $e\_d$ is the $d$-th standard basis vector, we get:


$$<\phi(a),\phi(b)> = \frac{\sigma(\gcd(a,b))}{\sigma(a) \sigma(b)}$$


which should be equal to:


$$\frac{1}{\sigma\left(\dfrac{ab}{\gcd(a,b)}\right)}$$


but I have no proof for this and have not seen this equality in any textbook. I am sure the proof is not difficult, but I can not come up with a proof for this at the moment, so any help is appreciated.



Can the Lagarias inequality be written as a "kernel inequality"?
There are several questions being asked here. I'll just deal with the easiest one, showing that $${\sigma(a)\sigma(b)\over\sigma(\gcd(a,b))}=\sigma\left({ab\over\gcd(a,b)}\right)$$


Let $p$ be a prime, let $p^c\|a$ (meaning $a$ is a multiple of $p^c$ but not of $p^{c+1}$), let $p^d\|b$, let $r$ be the larger, and $s$ the smaller, of $c$ and $d$. Note that $r+s=c+d$, $p^{r+s}\|ab$, $p^s\|\gcd(a,b)$, $p^r\|{ab\over\gcd(a,b)}$. Thus, the contribution of the prime $p$ to $\sigma\left({ab\over\gcd(a,b)}\right)$ is ${p^r-1\over p-1}$.


The contribution of $p$ to ${\sigma(a)\sigma(b)\over\sigma(\gcd(a,b))}$ is $${{p^c-1\over p-1}\cdot{p^d-1\over p-1}\over(p^s-1)/(p-1)}={p^r-1\over p-1}$$ and we're done.


