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Topic: Triangle inequality (Read 728 times) |
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NickH
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Triangle inequality
« on: Apr 23rd, 2003, 11:41am » |
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The sides of a triangle have lengths a, b, c. Show that:
3/2 <= a/(b + c) + b/(c + a) + c/(a + b) < 2.
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Nick's Mathematical Puzzles
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Mond
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This should work... the greater than applies to any 3 positive reals, but the lesser than inequality requires that they're sides of a triangle.
the greater than case :
a/(b+c) = (a+b+c)/(b+c) - 1 ....
this gives
(a+b+c) (1/(b+c) + 1/(c+a) + 1/(a+b)) - 3
=0.5((a+b)+(b+c)+(c+a)) (1/(b+c) + 1/(c+a) + 1/(a+b)) -3
since AM >= HM , (x+y+z)(1/x+1/y+1/z) >= 9
so the above is >= 9/2 - 3 (= 3/2)
the equality is when the triangle is equilateral
the lesser than :
since in a triangle, a+b > (a+b+c)/2 (because a+b>c)
a/(b+c) < a/s , where s = (a+b+c)/2
so the sum < (a+b+c)/s (=2)
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