Median inequality for three subharmonic functions

Definitions and mathematical statement

Let $u_1,u_2,u_3$ be subharmonic functions on $\mathbb C$ with $u_j(0)=0$. Subharmonic means upper semicontinuity together with the submean inequality on every disk; values $-\infty$ away from the origin are allowed. At each point arrange the three values in increasing order, giving $v_1(z)\leq v_2(z)\leq v_3(z)$. Thus $v_2$ is their median and $v_3$ their maximum; the median need not itself be subharmonic.

For $r>0$ put \[ I(r)=\int_0^{2\pi}v_2(re^{i\theta})\,d\theta,\qquad B(r)=\max_{|z|=r}v_3(z). \] The angular integral is unnormalized. The submean inequality at the origin implies $B(r)\geq0$. Define $R(r)=I(r)/B(r)$ when $B(r)>0$, and set $R(r)=0$ when $B(r)=0$. In the latter case the maximum principle forces all three functions to vanish on the disk, so $I(r)=0$ as well.

Must every such triple satisfy $\sup_{r>0}R(r)\geq0$? Does the stronger local assertion $\limsup_{r\downarrow0}R(r)\geq0$ always hold? These statements allow the ratio to remain negative at individual radii, or even approach zero from below; they do not demand a positive median mean at some specified radius. No growth restriction, harmonicity assumption, or common homogeneity is imposed on the three subharmonic functions.

Short English statement

Must the normalized circular mean of the median of three subharmonic functions be nonnegative in supremum, even arbitrarily near the origin?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3700039 (AMR-036-0039), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Eremenko - My favorite unsolved problems, Subharmonic inequality conjecture. Original source indexed by AMR-036-0039. https://www.math.purdue.edu/~eremenko/uns1.html.
• S3 A. Eremenko, Conjectured inequality for subharmonic functions. Author-maintained primary problem note. https://www.math.purdue.edu/~eremenko/dvi/subh1.pdf.