Fully powered mixed-ratio conjecture for arithmetic functions

Less than 1 year old · traced to 2026

Let $\varphi(n)$ , $\psi(n)$ and $\sigma(n)$ denote the Euler totient, Dedekind psi and sum-of-divisors functions, respectively. For integers $k\geq1$ and $n\geq2$ , form the three quantities $\varphi^k(n)(\psi^k(n)+\sigma^k(n))$ , $\psi^k(n)(\varphi^k(n)+\sigma^k(n))$ and $\sigma^k(n)(\varphi^k(n)+\psi^k(n))$ . Fully powered mixed-ratio conjecture. For every integers $k\geq1$ and $n\geq2$ , we have

\frac{\varphi^k(n)(\psi^k(n)+\sigma^k(n))}{\psi^k(n)(\varphi^k(n)+\sigma^k(n))+\sigma^k(n)(\varphi^k(n)+\psi^k(n))}+\frac{\psi^k(n)(\varphi^k(n)+\sigma^k(n))}{\varphi^k(n)(\psi^k(n)+\sigma^k(n))+\sigma^k(n)(\varphi^k(n)+\psi^k(n))}+\frac{\sigma^k(n)(\varphi^k(n)+\psi^k(n))}{\varphi^k(n)(\psi^k(n)+\sigma^k(n))+\psi^k(n)(\varphi^k(n)+\sigma^k(n))}\geq\frac{5(n-1)^{2k}+5(n^2-1)^k+2(n+1)^{2k}}{\big(3(n-1)^k+(n+1)^k\big)\big((n-1)^k+(n+1)^k\big)}.

This proposed inequality is not resolved in the supplied text.

References

Primary source

S. I. Dimitrov, “Lower bounds on expressions depending on the functions φ(n), ψ(n) and σ(n), III”, arXiv:2606.12484 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-written complete-proof attempt appeared, but no independent source has checked it, so the conjecture is not established.

S. I. Dimitrov’s June 2026 preprint records the inequality as Conjecture 9 for all $k \geq 1$ and $n \geq 2$ . The preprint supplies no proof or counterexample.

Posted attempt

A reader-written argument claims a complete proof: after setting $x=\varphi(n)^k$ , $y=\psi(n)^k$ , $z=\sigma(n)^k$ , it asserts $0<x\leq(n-1)^k<(n+1)^k\leq y\leq z$ , proves monotonicity of the resulting expression, and obtains the conjectured bound. The attempt has not been independently verified.

Current status (as of August 2026): The conjecture is recorded in a June 2026 preprint, while a complete proof has been posted in discussion but remains unverified; no verified proof or counterexample is known.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

Complete proof for every $n\ge2$ and $k\ge1$ .

Set

x=\varphi(n)^k,\quad y=\psi(n)^k,\quad z=\sigma(n)^k,\quad a=(n-1)^k,\quad b=(n+1)^k.

The Euler-product formulas imply $0<x\le a<b\le y\le z$ : indeed, $\varphi(n)\le n-1$ , $\psi(n)\ge n+1$ , and $\sigma(p^e)\ge p^e+p^{e-1}=\psi(p^e)$ .

Put

A=x(y+z),\quad B=y(x+z),\quad C=z(x+y),\qquad G=\frac A{B+C}+\frac B{A+C}+\frac C{A+B}.

By homogeneity let $r=x/y$ and $t=z/y$ , so $0<r\le a/b<1$ , $t\ge1$ . A direct common-denominator computation gives

G(r,t)-G(r,1) =\frac{r(t-1)P(r,t-1)} {(r+1)(3r+1)(rt+r+2t)(rt+2r+t)(2rt+r+t)},

where

\begin{aligned} P(r,v)={}&(2r^4+4r^3+7r^2+8r+3)v^2\\ &+(4r^4-5r^3+8r^2+19r+6)v\\ &+(1-r)(3r+1)(5r+3). \end{aligned}

Every denominator factor is positive. All coefficients of $P$ are nonnegative for $0<r\le1$ , since its linear coefficient equals

4r^4+5r^2(1-r)+3r^2+19r+6>0.

Thus $G(r,t)\ge G(r,1)$ . Moreover,

G(r,1)=\frac r{r+1}+\frac{2(r+1)}{3r+1},\qquad \frac{d}{dr}G(r,1) =\frac{(r-1)(5r+3)}{(r+1)^2(3r+1)^2}\le0.

Since $r\le a/b$ ,

G\ge G(a/b,1) =\frac{5a^2+5ab+2b^2}{(3a+b)(a+b)} = \frac{5(n-1)^{2k}+5(n^2-1)^k+2(n+1)^{2k}} {\big(3(n-1)^k+(n+1)^k\big)\big((n-1)^k+(n+1)^k\big)}.

This is the stated conjecture in full generality. Equality holds exactly when $n$ is prime.