Math Olympiad!!!?

This is an Asian Pacific Mathematics Olympiad problem. This question was posted on Dr.Math and they provided a very elegant proof. I want to see how you people might do it.





Let a, b and c be lengths of the sides of a triangle. Prove that the square root of (a + b - c) plus the square root of (b + c - a) plus the square root of (c + a - b) is equal to or less than the sum of the square roots of a, b and c. Determine when equality occurs.





Thanks for answering.

Math Olympiad!!!?
We wish to show that sqrt(a+b-c) + sqrt(b+c-a) + sqrt(c+a-b) %26lt;= sqrt(a) + sqrt(b) + sqrt(c). Equivalently we can show 2[sqrt(a+b-c) + sqrt(b+c-a) + sqrt(c+a-b)] %26lt;= 2[sqrt(a) + sqrt(b) + sqrt(c)].





Let's begin by showing sqrt(b+c-a) + sqrt(c+a-b) %26lt;= 2sqrt(c). Squaring both sides yields





(b+c-a) + (c+a-b) + 2sqrt((c+b-a)(c+a-b)) %26lt;= 4c


sqrt((c+b-a)(c+a-b)) %26lt;= c.





We square both sides again to get





(c+b-a)(c+a-b) %26lt;= c^2.


c^2 - (a-b)^2 %26lt;= c^2





which is clearly true since (a-b)^2 is nonnegative. Additionally equality is attained iff a=b. Thus, sqrt(b+c-a) + sqrt(c+a-b) %26lt;= 2sqrt(c). By symmetry we can generate two similar inequalities, sum them, and obtain the desired inequality. In the other two "subinequalities" equality will be attained iff b=c in one and iff a=c in the other. After summing them, equality will be attained iff the triangle is equilateral.





Very nice problem, btw!
Reply:how many times do i get this question asked. jeez do i have to answer this one again....

sweet pea