Bipartite connected graph proof

A friend presented me with a conjecture that seems to be true but neither of us can come up with a proof. Here's the problem:

Given a connected, bipartite graph with disjoint non-empty vertex sets U and V, such that |U|<|V|, all vertices are in either U or V, and there are no edges connecting two vertices within the 开发者_运维百科same set, then there exists at least one edge which connects vertices a∈U and b∈V such that degree(a)>degree(b)

It's trivial to prove that there is at least one vertex in U with degree higher than one in V, but to prove that a pair exists with an edge connecting them is stumping us.

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For any edge e=(a,b) with a∈U and b∈V, let w(e)=1/deg(b)-1/deg(a). For any vertex x, the sum of 1/deg(x) over all edges incident with x equals 1, because there are deg(x) such edges. Hence, the sum of w(e) over all edges e equals |V|-|U|. Since |V|-|U|>0, w(e)>0 for som edge e=(a,b), which means that deg(a)>deg(b).

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Prove it by contradiction, i.e. suppose that deg(a) ≤ deg(b) ∀(a,b)∈E, where E is the edgeset of the graph (with the convention that the first element is in U and the second in V).

For F⊆E, designate by V(F) the subset of V which is reachable through edgeset F, that is:

V(F) = { b | (a,b)∈F }

Now build an edgeset F as follows:

F = empty set
For a ∈ U:
    add any edge (a,b)∈E to F
Keep adding arbitrary edges (a,b)∈E to F until |V(F)| = |U|

The set V(F) obtained is connected to all nodes in U, hence by our assumption we must have

∑_a∈U deg(a) ≤ ∑_b∈V(F) deg(b)

However, since |U|=|V(F)| and |U|<|V| we know that there must be at least one "unreached" node v∈V\V(F), and since the graph is connected, deg(v)>0, so we obtain

∑_a∈U deg(a) < ∑_b∈V deg(b)

which is impossible; this should be an equality for a bipartite graph.