An element $F\in \mathbb{C}[[x,y]]$ defines a germ of plane curve.
We assume $F(0,0)=0$.
The multiplicity $mult$ of the germ is defined to be a minimal number $i$
such that $F\in m^i$ where $m=(x,y)$ is the maximal ideal in $\mathbb{C}[[x,y]]$.
Other standard invariants of the germ are Milnor number:


$$
\mu=\dim \mathbb{C}[[x,y]]/(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y})
$$


and delta invariant:


$$ \delta=\frac{\mu+r-1}{2}$$


where $r$ is number of branches of curve $F=0$ at $(0,0)$.


Question: What is the minimum of $\delta$ and $\mu$ among the germs
of given multiplicity?



Minimum of Milnor number for the curve singularities of fixed multiplicity 
As Roy remarked in his answer, the delta invariant of the germ of plane curve singularity $f(x,y)=0$ at $p$ is equal to


$\delta(f) = \sum \frac{m\_q(m\_q-1)}{2}$,


where the sum is extended over all the points $q$ which are "infinitely near" to $p$ and $m\_q$ denotes the multiplicity at $q$.


Then $\delta(f)$ is minimal among germs of a given multiplicity when there are no infinitely near points, in other words when the first blow-up of the germ is smooth. Of course there can be many analitically distinct germs satisfying this property: for instance, both the node and the ordinary cusp do the job among double points.


Now, since the Milnor number is equal to


$\mu(f)=2 \delta(f)-r(f)+1$,


where $r(f)$ is the number of branches, it follows that $\mu(f)$ is minimal among plane singularities of given multiplicity $n$ when $\delta(f)$ is minimal and the number of branches is maximal, in other words when $f=0$ is the "ordinary" $n$-ple point. 


As an example, for the ordinary double point (node) $y^2=x^3+x^2$ we have


$(\delta, \mu)=(1,1)$,


whereas the ordinary cusp $y^2=x^3$ satisfies


$(\delta, \mu)=(1,2)$.


Summing up, the ordinary $n$-ple point is the only germ of plane curve singularity which minimizes *both* $\delta$ and $\mu$, and the corresponding values are


$(\delta, \mu)= (\frac{n(n-1)}{2}, (n-1)^2)$.


