F.6 pure maths勁難！求高手解答，急！
(a) (i) Prove that x^2 + y^2 +z^2>=xy+yz+zx for any real numbers x,y and z.Let α ,β ,ɣ(measured in radians) be the interior angles of a triangle.
(ii) By using (i), prove that1/α^2 + 1/β
^2 +1/ɣ^2 >=27/π^2
(π即是Pi ,3.14159....我唔識打...)
(b) By using AM>=GM, prove that sinα+sinβ+sinɣ<=(3乘開方3)/2
HINT: cos((α-β)/2)<=1 本帖最後由 【YU】 於 7-12-2009 21:34 編輯
(x-y)^2 + (y-z)^2 + (z-x)^2 >= 0
x^2 - 2xy + y^2 + y^2 - 2yz + z^2 + z^2 - 2xz + x^2 >= 0
2x^2 + 2y^2 + 2z^2 >= 2xy + 2yz + 2zx
x^2 + y^2 + z^2 >= xy + yz + zx
let x = 1/a , y = 1/b , z = 1/c
1/a^2 + 1/b^2 + 1/c^2 >= 1/ab + 1/bc + 1/ca = (a+b+c)/abc
AM >= GM
(a+b+c)/3 >= 3rt abc
abc <= (a+b+c)^3/27
(a+b+c)/abc >= (a+b+c)/[(a+b+c)^3/27] = 27/(a+b+c)^2 = 27/pi^2
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