※ 引述《amy29585028 (阿金大人)》之銘言：
: http://ppt.cc/bkHg
: 高中學生拿來問的問題，應該是學校老師出的思考題，
: 想了一下不知道如何下手，懇請高手指點。
8bc 8ca 8ab
Let --- = a', --- = b', --- = c'
a^2 b^2 c^2
If exist such a,b,c that
1 1 1
------- + ------- + ------- ≧ 2
√(1+a') √(1+b') √(1+c')
1 1 1
Soppuse ------- = x, ------- = y, ------- = z,
√(1+a') √(1+b') √(1+c')
we have that x+y+z≧2, x,y,z<1.
Consider the triangle with side lengths x,y,z, the area of which D, the
radius of the circumcircle of which R,
16D^2
we have xyz=4RD and (x+y-z)(x-y+z)(-x+y+z) = -------
(x+y+z)
Now we have
(1-x^2)(1-y^2)(1-z^2) (2+2x)(2+2y)(2+2z)(2-2x)(2-2y)(2-2z)
--------------------- = ------------------------------------
x^2y^2z^2 64x^2y^2z^2
(3x+y+z)(x+3y+z)(x+y+3z)(x+y-z)(x-y+z)(-x+y+z)
≦ ----------------------------------------------
64x^2y^2z^2
(3x+y+z)(x+3y+z)(x+y+3z)
= ------------------------
64(x+y+z)R^2
125(x+y+z)^2
≦ ------------------------
1296R^2
(We have (x+y+z)^2≦27R^2 so)
125
≦ ------------------------
8
(1-x^2)(1-y^2)(1-z^2)
But --------------------- = a'b'c' = 512
x^2y^2z^2
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