Consider a graph $G$ with $g(n)$ edges. $G$ has average degree $2g(n)/n$ so there is a vertex $v$ with degree at most 
$$\delta = \left\lfloor \frac{2g(n)}{n}\right\rfloor.$$
Since by assumption $G-v$ has the property,
$g(n-1) \ge g(n)-\delta$.
Now solve this inequality for $g(n)$.

Write $2g(n)=an+b$ for $0\le b\le n-1$, then you can expand the floor and solve for $a$. Put that back into $an+b$ and you get (for $n\ge 3$) 
$$ g(n) \le \frac{n g(n-1)-b}{n-2},$$
which implies the given inequality.