1969 Canadian MO Problems/Problem 3

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Problem

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

Solution

By the Pythagorean Theorem and the trivial inequality, $2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0$.

Thus $2c^2\ge (a+b)^2.$ Since $a,b,c$ are all positive, taking a square root preserves the inequality and we have our result.

For equality to hold we must have $(a-b)^2 = 0$. In this case, we have an isosceles right triangle, and equality certainly holds for all such triangles.

1969 Canadian MO (Problems)
Preceded by
Problem 2
1 2 3 4 5 6 7 8 Followed by
Problem 4


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