MathLabs

Problem 5

Given n>4n>4 points in the plane, no three collinear, prove that at least (n−32)\binom{n-3}{2} convex quadrilaterals have their vertices among the given points.
Step 6 of 6: Factor and finish
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n(n−1)(n−2)−60(n−4)=(n−5)(n−6)(n+8)≥0n(n-1)(n-2)-60(n-4)=(n-5)(n-6)(n+8)\ge0
Detailed analysis

For n≥5n\ge5, the factorization is nonnegative: on 5≤n≤65\le n\le6 it is zero at the endpoints and nonnegative between, while for n≥6n\ge6 all factors have the required signs. Thus Q≥(n−32)Q\ge\binom{n-3}{2}.