Prove Triangle of Sides with Length...Date: 8/1/96 at 21:27:38 From: Anonymous Subject: Prove Triangle of Sides with Length... I have a problem for you from the Asian Pacific Mathematics Olympiad held on March 12, 1996: Question 5: Let a, b and c be the 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. Intuitively, equality occurs when the triangle is equilateral. My teacher has spent weeks on it but has been unable to reach a final proof. Would you please please please please please please please please please please please please please please please please please please please please pretty please help me to solve the problem. Date: 8/2/96 at 11:25:18 From: Doctor Anthony Subject: Re: Prove Triangle of Sides with Length... The information that a, b, c are the sides of a triangle has no significance beyond the fact that all the terms like a+b-c will be positive (two sides of a triangle are together greater than the third side). So we let a+b-c = p^2 b+c-a = q^2 c+a-b = r^2 From these equations 2a = r^2+p^2 2b = p^2+q^2 2c = q^2+r^2 So a = (r^2+p^2)/2 b = (p^2+q^2)/2 c = (q^2+r^2)/2 and we are asked to prove that: sqrt(a) + sqrt(b) + sqrt(c) >or= p + q + r ...(1) Now use the fact that (x^2+y^2)/2 > {(x+y)/2}^2 (See later for proof of this) Substituting for a, b, c in expression(1) we have: sqrt{(r^2+p^2)/2} + sqrt{(p^2+q^2)/2} + sqrt{(q^2+r^2)/2} >or= sqrt{(r+p)/2}^2 + sqrt{(p+q)/2}^2 + sqrt{(q+r)/2}^2 >or= (1/2){r+p + p+q + q+r} >or= (1/2){2p + 2q + 2r} >or= {p + q + r} And this is the required inequality, i.e. sqrt(a) + sqrt(b) + sqrt(c) >or= sqrt(a+b-c) + sqrt(b+c-a) + sqrt(c+a-b) NOTE: We must show that (x^2+y^2)/2 > {(x+y)/2}^2 as promised above. This inequality requires 2(x^2+y^2) > x^2 + 2xy + y^2 x^2 - 2xy + y^2 > 0 (x-y)^2 > 0 and this is always true. Hence the inequality is true. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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