1971 Canadian MO Problems/Problem 2
Let and be positive real numbers such that . Show that .
which is true since by AM-GM, we get: and we are given that , so
Let . Since we want to find the minimum of the function over the interval , we can take the derivative. Using the product rule, we get . Since we want this value to be zero, the numerator must be zero. But the only root of the equation is , and so plugging the answer back in we have Thus the minimum is and we are done.
|1971 Canadian MO (Problems)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •||Followed by|