Let $G$ be a simple graph on $k$ vertices without any cycle of length $3$ and every vertex in $G$ has degree at least $\frac{k+1}{4}$. Show $G$ is connected.
I assume that $G$ is not connected. Then there exists 2 vertices $u, v$ of $G$ such that there is no path from $u$ to $v$. Then I tried to find the number of vertices of $G$ which includes $u, v$ and their $\frac{k+1}{4}$ neighbors to get a contradiction. I don't know how to use the triangle-free assumption.
Any hint is highly appreciated.
Assume $G$ is not connected, and let $H$ be a component of $G$ with the fewest vertices $m$. Then $H$ contains at most half the vertices of $G$: i.e., $$ m \le \frac{k}{2}. $$ Because each vertex has degree at least $(k+1)/4$, the number of edges of $H$ satisfies $$ \frac{m}{2}\frac{k+1}{4} \le E(H). $$ However, Mantel's Theorem for triangle-free graphs states that $$ E(H) \le \frac{m^2}{4}. $$ Combining the two inequalities for $E(H)$ and dividing by $m/4$ yields $$ \frac{k+1}{2} \le m, $$ which contradicts the first inequality.