prove C-S ineq. Engel form

子路

2009-12-29 7:33 pm

by considering (ax-by)^2>=0, or ortherwise,
prove Cauchy-Schwarz's inequality in the Engel form
(generalized one)

圖片參考：http://i256.photobucket.com/albums/hh182/zilu_photo/sshot-2009-12-29-11-14-42.png

state the condition for equality to hold

please use elementary proof as possible
(yet I don't think you can rely much on other theorems)

更新1:

If you prove the first case, you can prove the inequality with self-generalizing property. ie. prove (a+b)^2/(x+y)>=a^2/x+b^2/y then sub b=b+c, y=y+z ... in this case, can you tell then condition for the equality to hold? (the generalized one)

更新2:

very good you have solved the problem I should have stated my aim clearly. actually, the self-generalizing property can be used as http://i256.photobucket.com/albums/hh182/zilu_photo/sshot-2009-12-30-02-21-42.png which gives a shorter proof

回答 (1)

用戶9112

2009-12-29 7:54 pm

✔ 最佳答案

(ax - by)^2 >= 0
a^2x^2 - 2abxy + b^2y^2 >= 0
a^2x^2/y - 2abx + b^2y >= 0
a^2(Σx^2/y) - 2ab(Σx) + b^2(Σy) >= 0 (subscripts omitted for clarity in Yahoo)
Consider this as a quadratic in a, the above inequality just means there is at most one solution for a, therefore discriminant <= 0
[2b(Σx)]^2 - 4(Σx^2/y)(b^2)(Σy) <= 0
[(Σx)]^2 - (Σx^2/y)(Σy) <= 0
[(Σx)]^2/(Σy) <= (Σx^2/y) thus proved
In order for the equality to hold,
Σ(ax - by)^2 = 0
That means for each x and y, x/y = b/a is a constant, i.e. each pair of x and y should be of the same ratio.

2009-12-29 13:38:52 補充：
Second proof by Induction:
http://img697.imageshack.us/img697/2073/induction.png

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