Let k be the number of elements common to all three sets, p be those in just A and B but not C, q in just B and C and r in just C and A, and let α be the number in just A, β in just B and γ in just C.
If that is not all clear, then the following diagram might help:
Now the three conditions are:
α+r=p+k, or p+k-r=α≥0;
β+p=q+k, or q+k-p=β≥0; and
γ+q=r+k, or r+k-q=γ≥0
We could use the three conditions to express any three of the variables in terms of the other four. It doesn’t matter which four we take as independent but a nice symmetric choice is to use p,q,r and k as we have done above to solve for α, β, and γ.
In terms of these we get
and our objective is to find extrema of
for positive integer values of k,p,q,r with p+k-r≥0, q+k-p≥0;, and r+k-q≥0; or equivalently r-p≤k, p-q≤k, and q-r≤k.
Equivalently, we could take k=1, look for rational values of p,q,r and rescale at the end to get integer solutions.
So we’re looking ar R=(1+p)/(1+r) with p,q,r≥0, and r-p≤1, p-q≤1, and q-r≤1.
Since r≤1+p, we get R≥(1+p)/(2+p)≥1/2
And since p-r=(p-q)+(q-r)≤2, we get p≤2+r so R≤(3+r)/(1+r)≤3
So R is never greater than 3, and the bound is attained when r=0 with q-r=1, and p-q=1.(which corresponds to the example cited earlier)