Suppose $G$ is not bipartite. A graph is bipartite if and only if it has no odd cycle, so choose a shortest odd cycle $C$ and write its length as $\ell$. Since there are no triangles, $\ell\ge5$.

The cycle has no chord. A chord would divide it into two shorter cycles whose lengths sum to $\ell+2$, an odd number. One of those shorter cycles would therefore be odd, contradicting the choice of $C$.

Every vertex outside $C$ has at most two neighbors on $C$. To prove this, suppose an outside vertex $x$ has at least three such neighbors. The consecutive neighbors divide $C$ into arcs, each of length at least two; adjacent neighbors would form a triangle with $x$. The arc lengths sum to the odd number $\ell$, so some arc has odd length $d$. There are at least two other arcs, each of length at least two, hence $d\le\ell-4$. This odd arc and the two edges from its endpoints to $x$ form an odd cycle of length $d+2<\ell$, again a contradiction.

Each vertex on $C$ has exactly two neighbors on $C$, and every other vertex has at most two. Counting incidences with the vertices of $C$ gives

$\ell\,\delta(G)\le\sum_{v\in V(C)}\deg(v)=\sum_{x\in V(G)}|N(x)\cap V(C)|\le2n.$

Therefore

$\delta(G)\le\frac{2n}{\ell}\le\frac{2n}{5},$

contradicting the hypothesis. Thus $G$ is bipartite.

For sharpness, take five disjoint vertex sets $V_0,\ldots,V_4$, each containing $t$ vertices. Join every vertex of $V_i$ to every vertex of $V_{i+1}$, with indices modulo five, and add no other edges. Each vertex has exactly $2t$ neighbors. No triangle is possible, since the five-cycle of parts has no three pairwise adjacent parts. Choosing one vertex from each part gives a five-cycle, so the graph is not bipartite.

This has $5t$ vertices and minimum degree $2t=2(5t)/5$, proving sharpness.

Source: Extremal Combinatorics, Po-Shen Loh, section 3, problem 2. This is the triangle-free case of the Andrasfai-Erdos-Sos theorem.