**Problem 3 of the Canadian Mathematical Olympiad 1969**

Let c be the length of the hypotenuse of a right angle triangle whose other two sides have lengths a and b. Prove that a + b ≤ c \sqrt{2}. When does the equality hold?

**Solution by Steve Dinh, a.k.a. Vo Duc Dien**

Applying the AM-GM inequality for any non-negative real numbers a and b, we have (a + b ) /2 ≥ \sqrt{ab},

and equality holds when a = b or

(a + b)² ≥ 4ab, or a² + b² + 2ab ≥ 4ab, or a² + b² ≥ 2ab (*)

Since c is the hypotenuse of a right triangle and a and b are the other two sides

c² = a² + b² and (*) becomes c² ≥ 2ab (**)

Now adding c² to the left and a² + b² to the right of (**), we have 2c² ≥ (a + b)², or a + b ≤ c \sqrt{2} equality holds when a = b