Young's Inequality & Its Consequences
Cambridge STEP 1 | 1987 | Question 6
Hello my dear friends of Math Games, welcome back to the 6th question of 1987’s Cambridge STEP Exam. As a reminder, these questions were designed to test out the most mathematically capable students hoping to read mathematics at the University of Cambridge.
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What good is a math problem if not to be solved with grit and wit? What follows is a complete solution, but a complete solution, read cold, teaches rather less than a partial attempt, wrestled with honestly.
Ergo, set aside a few minutes. Work through the steps. Let the difficulty be felt. The argument will mean considerably more once you have traced some part of it yourself.
Solution
Here we let
where g is the inverse function of f, so that
We are required to show that
Let’s interpret the integrals geometrically.
This integral F(s) represents the area under the curve y = f(x) from x = 0 to x = s.
Similarly,
The integral G(t) represents the area to the left of the curve x = g(y) from y = 0 to y = t.
Now let’s think about the rectangle with vertices
Its area is
The two regions represented by F(s) and G(t) together cover the whole rectangle, so their total area is at least the area of the rectangle.
Hence,
When does equality occur?
