Here's a proof, from Wikipedia, that $K_6$ has at least 2 monochromatic triangles. You may find that this method of proof generalizes to give you the lower bound you want.
"An alternate proof works by double counting. It goes as follows: Count the number of ordered triples of vertices $x, y, z$ such that the edge $(xy)$ is red and the edge $(yz)$ is blue. Firstly, any given vertex will be the middle of either $0\times5=0$ (all edges from the vertex are the same color), $1\times4=4$ (four are the same color, one is the other color), or $2\times3=6$ (three are the same color, two are the other color) such triples. Therefore there are at most $6\times6=36$ such triples. Secondly, for any non-monochromatic triangle $(xyz)$, there exist precisely two such triples. Therefore there are at most 18 non-monochromatic triangles. Therefore at least 2 of the 20 triangles in the $K_6$ are monochromatic."
On the other hand, if you want all the work done for you, there is an answer in A W Goodman, On sets of acquaintances and strangers at any party, Amer. Math. Monthly 66 (1959) 778-783, also in Frank Harary, The two-triangle case of the acquaintance graph, Math. Mag. 45 (1972) 130-135.
The algorithm is: pick an edge $x y_1$. Inductively find a $\Delta+1$ colouring $\phi$ of $G \setminus \{ x \to y_1 \}$.
For every vertex, there is a colour which is not used at that vertex, because the degree of the vertex is $\leq \Delta$ but we have $\Delta+1$ colours to choose from.
Let $c$ be a colour missing from vertex $x$, and $c_1$ missing from $y_1$.
Then if $c_1$ is missing at $x$, we are done: colour the edge $x y_1$ with colour $c_1$.
Otherwise, $c_1$ is not missing at $x$, so there is some $y_2$ in the neighbourhood $\Gamma$ of $x$ with $\phi(x \to y_2) = c_1$.
Repeat: given $y_1, \dots, y_k \in \Gamma(x)$ distinct, and distinct colours $c_1, \dots, c_k$ such that $c_i$ is missing at $y_i$ for all $i$, and $\phi(x y_i) = c_{i-1}$, do the following: if $c_k$ is missing at $x$, stop and recolour $x y_i$ with colour $c_i$. Otherwise, let $y_{k+1} \in \Gamma(x)$ be such that $\phi(x y_{k+1}) = c_k$, and let $c_{k+1}$ be a colour missing at $y_{k+1}$.
This process builds a list of distinct $y_i$ and $c_i$, so eventually it must terminate: it can only terminate if we find $c_{k+1}$ which has already appeared as an earlier $c_i$.
Wlog $c_1 = c_{k+1}$, because we can recolour $x y_j$ with colour $c_j$ for $1 \leq j \leq i-1$ and un-colour $x y_i$ itself.
Consider the subgraph of $G$ which is coloured only in colours $c_1, c$ (recalling that $c$ is the colour missing at $x$).
The maximum degree of this subgraph is $2$, so its components are paths and cycles; but $x, y_1, y_{k+1}$ all have degree $1$ in this subgraph and so must not lie in the same component.
If $x, y_1$ are disconnected in the subgraph, swap $c, c_1$ on the $c, c_1$-component of $y_1$, and let $\phi(x, y_1) = c$.
Otherwise, $x, y_{k+1}$ are disconnected in the subgraph, so swap $c$ and $c_1$ on the $c, c_1$-component of $y_{k+1}$, and recolour $x y_{k+1}$ by $c$, $x y_i$ by $c_i$ for each $i \in [1, k]$.
Best Answer
I'm assuming that by "vertex coloring" you are not referring to proper vertex coloring, since every proper vertex coloring has zero monochromatic edges.
I'll prove the claim using the Probabilistic method.
Given a graph $G=(V,E)$, and k colors, I'll define random coloring as a function $C:V\to[k]$ (where $[k]:=\{1,2....,k\}$) as follow: $$\forall v\in V, i\in[k]\ \ \ \ \mathbb {P}(c(v)=i)=\frac{1}{k}$$ Proposition $$\forall e\in E\ \ \ \mathbb {P}(e\text{ is monochromatic})=\frac{1}{k}$$ The proposition implies the expected number of monochromatic edges for such random coloring is $\frac{\vert E\vert}{k}$.
Therefore, at least one random coloring induce no more then $\frac{\vert E\vert}{k}$ monochromatic edges, otherwise, the expectation value would be higher.
The proposition proof is quite simple: $$\forall i\in [k],v,u\in V\ \ \ \ \mathbb P(c(u)=i\land c(v)=i) = \frac{1}{k^2}\implies\mathbb P(c(u)=c(v))=\frac{1}{k}$$ Where the last transition is true because there are k colors.