There is no non-constant positive concave function on all of $\mathbb{R}$.
If $f$ is a concave function, then the difference ratio $ \frac{f(x)-f(y)}{x-y} $ is a non-increasing function of both $x$ and $y$.
Suppose there are $x$ and $y$ so that $f(x) and let $M=\dfrac{f(x)-f(y)}{x-y}$.
If $x, then $M>0$ and for all $z< x$, we have that $ \frac{f(z)-f(y)}{z-y}\ge\frac{f(x)-f(y)}{x-y}=M\tag{1} $ therefore, $ f(z)\le f(y)+(z-y)M\tag{2} $ and $(2)$ says that for all $z< y-\frac{f(y)}{M}$, we have $f(z)<0$.
If $x>y$, then $M<0$ and for all $z>x$, we have that $ \frac{f(z)-f(y)}{z-y}\le\frac{f(x)-f(y)}{x-y}=M\tag{3} $ therefore, $ f(z)\le f(y)+(z-y)M\tag{4} $ and $(4)$ says that for all $z>y-\frac{f(y)}{M}$, we have $f(z)<0$.
Thus no matter whether $x>y$ or $x, we have that $f$ cannot stay positive.
Superharmonic Functions
A superharmonic function is one whose Laplacian is non-positive everywhere. I believe this is what you mean by "the Laplacian case".
Since the case of harmonic functions can be handled using Harnack's Inequality, let's consider the case of strictly superharmonic functions; that is, ones for which $\int_{\mathbb{R}^n}\Delta f(y)\,\mathrm{d}y\lt0$.
Let $B(r,x)=\{y\in\mathbb{R}^n:|y-x|\le r\}$. Then, by the Divergence Theorem, $ \int_{B(r,x)}\Delta f(y)\,\mathrm{d}y=\int_{\partial B(r,x)}\nabla f(y)\cdot\vec{n}\,\mathrm{d}\sigma(y)\tag{5} $ The right hand side of $(5)$ is the measure of $\partial B(r,x)$ times the rate of change of $m_f(r,x)$, the mean value of $f$ over $\partial B(r,x)$. The left hand side of $(5)$ is strictly negative past a certain $r_0$. Rewriting this, we get for $r\ge r_0$, $ \omega_{n-1}r^{n-1}\frac{\partial m_f(r,x)}{\partial r}\le C\lt0\tag{6} $ which leads to $ \frac{\partial m_f(r,x)}{\partial r}\le\frac{C}{\omega_{n-1}r^{n-1}}\tag{7} $ For $n\le2$, $(7)$ says that there can be no lower bound for $m_f(r,x)$, which means there can be no lower bound for $f$.
According to item $8)$ here, a superharmonic function on the whole of $\mathbb{R}^2$ that is bounded below is constant. However, the same is not true on $\mathbb{R}^n$ for $n\ge3$. This agrees with $(7)$.