A question about a part of a proof of Picard's little theorem. Picard's little theorem states that if $f$ is a non-constant entire function, then $\mathbb{C}\setminus f(\mathbb{C})$ contains at most one point.
Now in the proof we assume that $\alpha,\beta\in \mathbb{C}\setminus f(\mathbb{C})$ and we define two harmonic functions $h:=\log|f-\alpha|$ and $k:=\log|f-\beta|$ on $\mathbb{C}$. Why does it then follow that $$|h^{+}-k^+|\leq |\alpha-\beta|$$ and $$\max(h,k)\geq \log(|\alpha-\beta|/2)$$
 A: $\bullet$ If $| f - \alpha|, | f - \beta|$ are both $\leq 1$, then $h^+ - k^+ = 0$, then the first inequality is trivial. 
$\bullet$ If $| f - \alpha|\leq 1$, and  $| f - \beta| > 1$, then 
$$
\vert h^+ - k^+ \vert = \log \vert f - \beta \vert = \log \vert f -\alpha + \alpha- \beta \vert \leq  \log\big( \vert f -\alpha \vert + \vert \alpha- \beta \vert \big) \leq \log\big( 1 + \vert \alpha- \beta \vert  \big)
$$
which is $\leq \vert \alpha- \beta \vert$.
$\bullet$ If $| f - \alpha| \geq | f - \beta|\geq 1$, then 
$$
\vert h^+ - k^+ \vert =  \log  \vert\frac{ f - \alpha }{ f - \beta}\vert = \log\Big( \vert 1 + \frac{\beta- \alpha}{f -\beta} \vert \Big)\leq \log\Big( 1 + \vert \alpha - \beta\vert \Big) \leq \vert \alpha- \beta \vert
$$
This gives us the first inequality.
the second inequality is equivalent to 
$$\max \Big\{ \vert f - \alpha\vert,  \vert f - \beta\vert \Big\} \geq  \frac{\vert \alpha - \beta\vert }{2} $$
while it follows from the basic triangle-inequality that
$$2\max \Big\{ \vert f - \alpha\vert,  \vert f - \beta\vert \Big\}\geq  \vert f - \alpha\vert +  \vert f - \beta\vert \geq \vert \alpha - \beta\vert \,.
$$
The proof is finished now. 
