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Write $r=|V_0|$, $s=|V_1|$, and $e=|E(G)|=\alpha |X||Y|$. We construct a random graph homomorphism $\Psi :T\to G$. Choose a root of the tree. Map it according to the degree-biased vertex distribution on the corresponding side of the bipartite graph; after mapping any vertex, map each child independently and uniformly to a neighbour of its parent's image.

Every oriented tree edge is then mapped uniformly onto the $e$ edges of $G$. Every tree vertex in $V_0$ has the degree-biased marginal on $X$, of entropy $H_X$, and every vertex in $V_1$ has the analogous marginal of entropy $H_Y$. Repeated use of the chain rule for information entropy along the rooted tree givesSince a $k$-vertex tree has $k-1$ edges,The maximum entropy distribution on a finite set gives $H_X\le \log |X|$ and $H_Y\le \log |Y|$. HenceIf $N$ is the number of bipartition-respecting graph homomorphisms $T\to G$, the support of $\Psi$ has size $N$, so the maximum entropy distribution on a finite set also gives $H(\Psi )\le \log N$. ThusThere are $|X{|}^r|Y{|}^s$ bipartition-respecting maps in total, so a uniformly chosen one is a graph homomorphism with probability at least ${\alpha }^{k-1}$. This proves the Sidorenko inequality for trees.