# prove C-S ineq. Engel form

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_phot...

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)

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)

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_phot...

which gives a shorter proof

### 1 Answer

- 自由自在Lv 71 decade agoFavorite Answer
(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:

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