Is there a harmonic function in the whole plane that is positive everywhere? This is one of the past qualifying exam problems that I was working on. 
I know that, when we let $z=x+iy$, ${|z|}^2=x^2+y^2$ is not harmonic. I do not know where to start to prove that there is no harmonic function that is positive everywhere.
Any help or ideas idea will be really appreciated. 
Thank you in advance. 
 A: Let $f(z)$ be entire and positive. Consider 
$$h(z) = e^{-f(z)}$$
If $\Re f(z) > 0$, we have $-\Re f(z) < 0$, so $h(z)$ is bounded and entire. What can you conclude about $h(z)$, and hence about $f(z)$?

To go into a little more detail, note that
$$|h(z)| = e^{-\Re f(z)} < e^{0} < 1$$
A: Continuing,a non-constant harmonic function is the real part of a non-constant entire function.so the real part must be positive.Little Picard Theorem: If a function  is entire and non-constant, then the set of values that f(z) assumes is either the whole complex plane or the plane minus a single point. so we get a contradiction
A: MORE GENERAL AND SIMPLE ANSWER

Liouville Type (Theorem's)
  Suppose $u:\mathbb R^d\to \mathbb R$ be a nonegaitve harmonic function 
  Then, $u$ is constant.

Proof
$u\ge 0$ then $\inf_{\mathbb R^d} u<\infty$. Hence we set 
$$v= u-\inf_{\mathbb R^d} u$$
and 
\begin{split} \begin{cases}\Delta v=0\\
v\ge 0\\
\inf_{\mathbb R^d} v =0\end{cases}\end{split}
Let $\varepsilon>0,$ and $y\in\mathbb R^d$  then there exists $x_\varepsilon$ such that,
$$ v(x_\varepsilon)\le \inf_{\mathbb R^d} v+\varepsilon$$
Let $R>\max(|x_\varepsilon|,|y|)+1$ therefore, $x_\varepsilon,y\in B_R(0)$ and 
$$ B_R(y)\subset B_{3R}(x_\varepsilon) $$
By Mean value property (The mean value is true in any dimension for harmonics functions),
\begin{split} v(y) &=& \frac{1}{|B_R(y)|}\int_{B_R(y)} v(z) dz\\ &=& \frac{3^d}{|B_{3R}(x_\varepsilon)|}\int_{B_R(y)} v(z) dz \\&\le&\frac{3^d}{|B_{3R}(x_\varepsilon)|}\int_{B_{3R}(x_\varepsilon)} v(z) dz \\&= & 3^d v(x_\varepsilon) \end{split}
This leads to,
$$  v(y)\le 3^d v(x_\varepsilon)\le 3^d(\inf_{\mathbb R^d} v+\varepsilon)~~\forall y\in  \mathbb R^d$$
That is 
$$  \sup_{\mathbb R^d} v < 
  3^d \varepsilon $$
Since $\inf_{\mathbb R^d} v=0$ ~~ but $\varepsilon>0$ was arbitrarily chosen, letting $\varepsilon \to 0$ we get 
$$ 0\le \sup_{\mathbb R^d} v  \le 0$$
i.e $v = 0$ or $u= \inf_{\mathbb R^d} u $
