TITLE: Maximal number of connected components of complement to an affine plane real algebraic curve
QUESTION [7 upvotes]: Let $X$ be a (singular, reducible) affine plane real algebraic curve of degree $d$.
How we can estimate maximal number of connected components of it's complement in $R^2$ in terms of degree?

REPLY [2 votes]: We can also prove that the maximal number of components for reducible nonsingular curve is $\frac{n^2+n+2}{2}$ by the following computation.
Let $m$ be a partition of $n$, where $m_i$ is the degree of $i$-th irreducible component of our curve.
Denote $s=\{i|m_i=1\}$, $t=\{i|m_i>1\}$.
Then, using Harnack inequality and Besout's theorem(for intersections with infinite line and for intersections between irreducible components) we can write that the number of connected components is lesser than
$$
\max_{m\vdash n}(\sum_{i}(\frac{(m_i-1)(m_i-2)}{2}+1)+1+n-s+\sum_{i\neq j}m_im_j)=$$
$$=\frac{n^2}{2}-n+1+\max_{m\vdash n}(2t+s))\leq $$
$$\leq \frac{n^2}{2}-n+1+ \max_{0\leq 2x+y\leq n,0\leq x,0\leq y}(2x+y)=\frac{n^2+n+2}{2}
$$ 
Therefore the only question we have is the following:
why do Harnack bound (in terms of degree) holds for singular curves as well as for nonsingular?
The proof could be obtained using First Harnack Theorem.