= Solution
\b[The conclusion holds in dimension two.] Shift and change sign to obtain a nonnegative <harmonic function> $v$. Its circular mean satisfies
$$
m(r)=\frac1{2\pi}\int_0^{2\pi}v(r\cos\theta,r\sin\theta)\,d\theta,\qquad
m''+r^{-1}m'=0.
$$
Here the angular second derivative in the polar-coordinate <Laplacian> integrates to zero. Thus $m(r)=a\log r+b$; nonnegativity for every $r>0$ forces $a=0$. For $|x|=r$, the <mean value property for harmonic functions> and nonnegativity give
$$
v(x)\leq\frac4{\pi r^2}\int_{r/2<|z|<3r/2}v(z)\,dz=8b.
$$
The <removable singularity for a bounded harmonic function> fills in the origin, and the <harmonic Liouville theorem> makes the extension constant.
\b[The conclusion fails for $n\geq3$:] $\boxed{u(x)=|x|^{2-n}}$ is positive and nonconstant, and its radial <Laplacian> $u''+(n-1)u'/r$ vanishes away from zero. For $n=1$, $\boxed{u(x)=|x|}$ is a nonconstant function bounded below whose second derivative vanishes on both components of the punctured line.
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