Let varphi(n), psi(n) and sigma(n) denote the Euler totient, Dedekind psi and sum-of-divisors functions, respectively. For integers k≥1 and n≥2, use varphi^k(n), psi^k(n) and sigma^k(n) for their kth powers. Power-ratio conjecture. For every integers k≥1 and n≥2, we have fracvarphi^k(n)psi^k(n)+sigma^k(n)+fracpsi^k(n)varphi^k(n)+sigma^k(n)+fracsigma^k(n)varphi^k(n)+psi^k(n)≥frac4(n+1)^2k+(n^2-1)^k+(n-1)^2k2 big((n+1)^2k+(n^2-1)^k big). The source verifies this conjecture for k=2 and k=3, while the general assertion remains open.

Power-ratio conjecture for Euler's, Dedekind's and sum-of-divisors functions

From papers

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 $n\geq2$ , use $\varphi^k(n)$ , $\psi^k(n)$ and $\sigma^k(n)$ for their $k$ th powers. Power-ratio conjecture. For every integers $k\geq1$ and $n\geq2$ , we have

\frac{\varphi^k(n)}{\psi^k(n)+\sigma^k(n)}+\frac{\psi^k(n)}{\varphi^k(n)+\sigma^k(n)}+\frac{\sigma^k(n)}{\varphi^k(n)+\psi^k(n)}\geq\frac{4(n+1)^{2k}+(n^2-1)^k+(n-1)^{2k}}{2\big((n+1)^{2k}+(n^2-1)^k\big)}.

The source verifies this conjecture for $k=2$ and $k=3$ , while the general assertion remains open.

Progress summary

Partially solved

A 2026 paper proves the inequality for the second and third powers, but the conjecture for all powers remains open.

The conjecture asserts the stated lower bound for every integer $k\geq1$ and $n\geq2$ . A June 2026 preprint records it as Conjecture 3 and gives no general proof or counterexample.

Known results

The conjecture is established for $k=2$ and $k=3$ by the method used for the paper's Theorem 1.

Current status (as of August 2026): The cases $k=2$ and $k=3$ are settled, while the assertion for general $k\geq1$ remains open.

Sources

arxiv.org

Sources & referencesView supporting material

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).

Additional references

3 papers in this index state this conjecture (2017–2026). The statement above is taken from the most recent of them; the others are arXiv:2401.09497, arXiv:1712.08666.

1 week ago

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

Put

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 give

0<x\le a<b\le y\le z.

Indeed, $\varphi(n)\le n-1$ , $\psi(n)=n\prod_{p\mid n}(1+1/p)\ge n+1$ , and $\sigma(p^e)\ge p^e+p^{e-1}=\psi(p^e)$ for every prime power.

Define

S(x,y,z)=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}.

For fixed $x,y$ ,

\frac{\partial S}{\partial z} =\frac1{x+y}-\frac{x}{(y+z)^2}-\frac{y}{(x+z)^2}.

This derivative increases with $z$ , and at $z=y$ it is

x\left(\frac1{(x+y)^2}-\frac1{4y^2}\right)\ge0.

Hence $S(x,y,z)\ge S(x,y,y)$ . Write $r=x/y$ . Then

S(x,y,y)=\frac r2+\frac2{1+r},\qquad \frac{d}{dr}\left(\frac r2+\frac2{1+r}\right) =\frac12-\frac2{(1+r)^2}\le0

for $0<r\le1$ . Since $r\le a/b$ ,

S(x,y,z)\ge \frac{a^2+ab+4b^2}{2b(a+b)} = \frac{4(n+1)^{2k}+(n^2-1)^k+(n-1)^{2k}} {2\big((n+1)^{2k}+(n^2-1)^k\big)}.

This is precisely the conjectured inequality. Equality holds exactly when $n$ is prime.

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