F.6 pure maths勁難！求高手解答，急！

麥海智偷拍

(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

[S]【YU】

(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