Triangle ProofsDate: 06/28/98 at 04:03:32 From: Dr. H. M. Behl Subject: Problem with a question Dear Dr. Math, Could you please help me with these two questions? I cannot figure out how to start doing the problems. 1. If the sides of a triangle are a, b, c, then prove that (a+b+c)^3 >= 27(b+c-a)(c+a-b)(a+b-c) 2. If a>0, b>0, c>0 then prove that a/b+c + b/c+a + c/a+b >= 3/2 Thanking you and waiting for your reply, Rohit Behl Date: 06/28/98 at 08:57:36 From: Doctor Anthony Subject: Re: Problem with a question >(1) If the sides of a triangle are a, b, c, then prove that (a+b+c)^3 >= 27(b+c-a)(c+a-b)(a+b-c) Use the fact that the arithmetic mean of three positive numbers is greater than the geometric mean. So x+y+z ------- > (xyz)^(1/3) 3 (x+y+z)^3 > 27(xyz) ........ (1) Now let x = a+b-c y = b+c-a z = c+a-b (since a,b,c are sides of a triangle, and two sides are always greater than the third side, x, y, z are positive) x+y+z = 2a+2b+2c-a-b-c = a+b+c and we get, substituting into (1) (a+b+c)^3 > 27(a+b-c)(b+c-a)(c+a-b) >2. If a>0, b>0, c>0 then prove that a/b+c + b/c+a + c/a+b >= 3/2 To prove this we need first to show that the least value of the lefthand side occurs when a/(b+c) = b/(c+a) = c/(a+b). If a+b+c = constant and if we assume that c is fixed, then a+b is fixed but we vary a and b to minimize the lefthand side, so c/(a+b) = constant and we consider the sum of the other two fractions. a/(b+c) + b/(c+a) a(c+a) + b(b+c) a^2 + b^2 + c(a+b) = -------------- = ------------------- (b+c)(c+a) (b+c)(c+a) (a+b)^2 - 2ab + c(a+b) = ----------------------- remember (b+c)(c+a) a+b is fixed, the top line is a minimum if 2ab is a maximum, and the bottom line is a maximum if (b+c)(c+a) is a maximum, i.e. if the two factors are equal. Both these conditions require a = b. With a = b, if we now allow c to vary, a similar argument leads to c = b, and so for the least value on the lefthand side we require a = b = c. Then with a = b = c we obtain the value of the lefthand side: 1/2 + 1/2 + 1/2 = 3/2 and this is the minimum value. It follows that a/(b+c) + b/(c+a) + c/(a+b) >= 3/2 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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