Re: Two algebraic inequalies.....
- From: Valeriu Anisiu <vanisiu@xxxxxxxxxxx>
- Date: Tue, 06 Dec 2005 05:55:53 EST
> If -infty < a =< b =< c =< d < infty, it's true
> that
> ===================================================
>
>
>
> 16(c-b)(c-a)(b-a)(d-a)(d-b)(d-c)
> )(b-a)(d-a)(d-b)(d-c) =<
> (1)
> =< (c-b)(d-a)(b+c-a-d)^4
> b+c-a-d)^4 +(b-a)(d-c)(a+b-c-d)^4
>
> ===================================================
> and
> =========================================
>
>
>
>
>
>
>
> 8(c+a-b-d)^2(c-a)(d-b)*sqrt((c-b)(b-a)(d-a)(d-c))
> -c)) =<
> (2)
> =< (c-b)(d-a)(b+c-a-d)^4
> a)(b+c-a-d)^4 +(b-a)(d-c)(a+b-c-d)^4 ?
>
> =========================================
> (What about equality cases in (1) , (2) ? )
>
The solutions are very simple using Maple, introducing
x= b-a >= 0 etc:
[1]
> u := 16*(c-b)*(c-a)*(b-a)*(d-a)*(d-b)*(d-c):
> v := (c-b)*(d-a)*(b+c-a-d)^4+(b-a)*(d-c)*(a+b-c-d)^4:
> factor(subs([b=a+x,c=a+x+y,d=a+x+y+z],v-u));
(y+z)*(x+z)^4*(x+y)
So, equality holds iff two of the numbers a,b,c are equal.
[2]
> u := 64*( (c+a-b-d)^2*(c-a)*(d-b) )^2 *(c-b)*(b-a)*(d-a)*(d-c):
> v:= ( (c-b)*(d-a)*(b+c-a-d)^4 +(b-a)*(d-c)*(a+b-c-d)^4 )^2:
> factor(subs([b=a+x,c=a+x+y,d=a+x+y+z],v-u));
(y+z)^2*(x+y)^2*(x^4-16*x*y^2*z-16*x^2*y*z+4*x^3*z-16*x*y*z^2+6*x^2*z^2+4*x*z^3+z^4)^2
So the inequality holds.
The equality is more difficult; it holds when x=y=0 or y=z=0 or
x^4-16*x*y^2*z-16*x^2*y*z+4*x^3*z-16*x*y*z^2+6*x^2*z^2+4*x*z^3+z^4=0
(for example, x = 6^(1/2) +2^(1/2) - 1, y=1, z=1)
V. Anisiu
.
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