Theorem: If $G$ is a graph having $p$ vertices and $\delta(G)\ge\frac{p-1}{2}$, then the edge connectivity of $G$ is $\delta(G)$.
$\delta\left(G\right)$ denotes the minimum degree of $G$.
Source: Graph Theory; Bondy & Murty
Let $\delta (G) = d$ and suppose we can delete $r<d$ edges (color them red) so that $G$ becomes unconnected with components $A,B$. We can assume that $|A|\geq |B|=:b$ so $b\leq {p\over 2}$
Redelete this red edges for a time.
Take any $v$ in $B$. Vertex $v$ is connected with at most $b-1$ vertices in $B$, so it must be connected with at least $d_v-(b-1)$ vertices in $A$. Say $B= \{v_1,v_2,...v_b\}$ so we have at least $$\Big(d_1-(b-1)\Big)+\Big(d_2-(b-1)\Big)+...+\Big(d_b-(b-1)\Big)$$ edges from $B$ to $A$ so they are all red. Thus we have $$r\geq b\cdot d -b(b-1)$$ Since we supposed $r<d$ we have now $$d>bd-b(b-1)\implies b>d \;\;\;{\rm and}\;\;\;b\ne 1$$
So we have $${p\over 2}\geq b\geq d+1 \geq{p-1\over 2}+1={p+1\over 2}$$ a contradiction.