Let $$H_n=1+\frac{1}{2}+\cdots+\frac{1}{n},$$ the nth harmonic number and $$\sigma(n)=\sum_{d\mid n}d,$$ the sum of divisor function, for example $\sigma(6)=12$.
I believe that this could be a nice problem in analysis.
Question. For which integer $M\geq 3$ can you claim easily, without assuming Riemann Hypothesis, that previous arithmetical functions satisfies $$\sigma(n)\leq H_n+M^{H_n}\cdot\frac{\log H_n}{\log M},$$ for all $n\geq 1$?
In a phrase, what is the minimum integer $M\geq e$ that satisfies a modified Lagarias' statement, and it is easily to prove, unconditionally, with analysis?
Thanks in advance.
Looking at the formulation you want with $M,$ $M=3$ works, indeed any real $M > e$ works.
Lemma 3.1 in Lagarias is, for $n \geq 3,$
$$ e^\gamma n \log \log n \leq \exp (H_n) \log (H_n) $$
Formula (2.2), Robin showed, unconditionally, for $n \geq 3,$ $$ \sigma(n) < e^\gamma n \log \log n + 0.6482 \frac{n}{\log \log n}. $$
Without using the extra $H_n$ at all, we get $$ \sigma(n) < \exp (H_n) \log (H_n) + 0.6482 \frac{n}{\log \log n}. $$
If we include the $H_n,$ formula (3.7) is, this time for $n \geq 20,$ $$ H_n + \exp (H_n) \log (H_n) \leq e^\gamma n \log \log n + \frac{7n}{ \log n}.$$