1969 Canadian MO Problems/Problem 3

Revision as of 09:21, 28 July 2006 by JBL (talk | contribs) (Solution: SOLVE INEQUALITIES IN THE PROPER DIRECTION!!!!!!!)


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


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

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

The equality condition is clearly that $(a-b)^2 = 0$ -- the isosceles right triangle.

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