Inequalities are fun. Today in analysis we proved Young’s Inequality in an exceptionally clever way. I do not wish to deprive the reader of this beautiful proof, so I shall recreate it here:
Theorem: If p>=1, and 1/p+1/p’=1 (we don’t worry too much about infinity in this class), and s and t are >= 0, then st 0. Let (s,t) be a point in the xy-plane. The area of the rectangle with vertices (0,0), (s,0), (s,t), (0,t) is st. Now consider the integral from 0 to s of x^(alpha) dx and the integral from 0 to t of y^(1/alpha) dy. The sum of these integrals is greater than or equal to st. Now, when we evaluate these integrals and sum them, we get st =0 wasn’t elegant enough. Unfortunately, he couldn’t remember it, so I tried to help him out, but I needed a bit of time to think about it, and I wasn’t given it. After class, I saw one of the students trying to figure it out on the chalkboard, so I told him how it worked. Then he showed me a neat proof of the theorem that states that if a point P is on the circumcircle of ABC, then the feet of the perpendiculars from P to AB, BC, and CA are collinear.