Inequalities :::::::::::::::

[匿名]

2012-08-07 7:28 am

Let a,b and c denote the lengths of the three sides opposite to the three angles
A,B and C of a triangle.
Show that

π/3 =< (aA+bB+cC)/(a+b+c) =< π/2

回答 (1)

用戶25975

2012-08-07 8:08 pm

✔ 最佳答案

(a - b)(A - B) >= 0

(b - c)(B - C) >= 0

(c - a)(C - A) >= 0

Adding them together:

(a - b)(A - B) + (b - c)(B - C) + (c - a)(C - A) >= 0

2 (aA + bB + cC) - (b + c)A - (c + a)B - (a + b)C >= 0

2 (aA + bB + cC) >= (b + c)A + (c + a)B + (a + b)C

Adding aA + bB + cC to both sides:

3 (aA + bB + cC) >= (a + b + c)A + (a + b + c)B + (a + b + c)C

3 (aA + bB + cC) >= (a + b + c)(A + B + C) = π(a + b + c)

(aA + bB + cC)/(a + b + c) >= π/3

On the other hand:

(a + b + c)A >= 2aA

(a + b + c)B >= 2bB

(a + b + c)C >= 2cC

(a + b + c)(A + B + C) >= 2(aA + bB + cC)

π(a + b + c) >= 2(aA + bB + cC)

(aA + bB + cC)/(a + b + c) <= π/2

參考: 原創答案

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