We start by proving an analogous bound for the \(3\)-vertex path \(P_3\text{.}\) The number of homomorphisms from \(P_3\) to \(G\) is precisely
\begin{equation*}
\sum_{v\in V(G)}d(v)^2
\end{equation*}
\begin{equation*}
\frac{1}{|V(G)|}\left(\sum_{v\in V(G)}d(v)\right)^2 = \frac{4|E(G)|^2}{|V(G)|}\text{.}
\end{equation*}
Thus, dividing both sides by
\(|V(G)|^3\) and recalling
Example 4.2.4 yields
\begin{equation}
t(P_3,G)\geq t(K_2,G)^2\text{.}\tag{4.2.1}
\end{equation}
We now apply a similar argument to turn this into a lower bound on \(t(C_4,G)\text{.}\) The number of homomorphisms from \(C_4\) to \(G\) is precisely
\begin{equation*}
\sum_{(u,v)\in V(G)^2}|N(u)\cap N(v)|^2
\end{equation*}
\begin{equation*}
\frac{1}{|V(G)|^2}\left(\sum_{(u,v)\in V(G)^2}|N(u)\cap N(v)|\right)^2\text{.}
\end{equation*}
Now, the key observation here is that
\(\sum_{(u,v)\in V(G)^2}|N(u)\cap N(v)|\) is, in fact, also equal to the number of homomorphisms from
\(P_3\) to
\(G\text{.}\) Therefore, dividing both sides by
\(|V(G)|^4\) and applying
(4.2.1) gives us
\begin{equation*}
t(C_4,G)\geq \frac{1}{|V(G)|^6}\left(t(P_3,G)|V(G)|^3\right)^2 = t(P_3,G)^2 \geq t(K_2,G)^4\text{.}
\end{equation*}
This completes the proof.