Assuming x, y, z, and t are positive, we list and prove some inequalities that are used in this paper to obtain error bounds:
In the next five inequalities let ( a = (x+y)/2 ) and ( g = xy ). Inequalities become equalities in (
), (), and () iff x = y.
or alternatively,
Finally we have
where ( a = (x+y+z)/3 ) and ( b = 3(xy+xz+yz)/2 ), and
where ( g = (xyz)^1/3 ) and ( 3h^-1 = x^-1+y^-1+z^-1 ).
To prove () we write
[
1t - 1t+x = t+x- tt(t+x) = xt(t+x)(t+x+t) ]
and replace the last denominator factor by either ( 2t ) or ( 2t+x ). Interchange of t and x leads from () to (). To prove () let (y = 1+x/t ) and write
[
t^3/2(t+x)x(1t^3/2-1(t+x)^3/2)
= y^2y^2-1(1-1y^3) = 1+1y(y+1), ]
which increases from 1 to 
as t increases from 0 to 
and y decreases from 
to 1. Interchange of t and x leads from () to ().
If the left side of () is put over a common denominator, it suffices to observe that
The left inequality is enough to prove the left inequality in (). To prove the right inequality in (), we define
[
(t) = ((t+x)(t+y) - xy)/t ]
and note that ( (t) ) tends to 
as ( t 0 ) and to 1 as ( t ). Differentiation shows that 
decreases monotonically, because
[
t^2(t+x)(t+y)' = -(ta+g^2) + [(ta+g^2)^2 - t^2(a^2-g^2)]^1/2
0,
]
with equality iff ( x=y ). Because of (), () implies
().
Equation () is proved by solving for 
and using (). Likewise, () is proved by solving for
[
= (t) + tt+z
]
and using the result just established that ( 1 (t) a/g ).
To prove () we use Maclaurin's inequality [17, Thm. 52,] to find that
[
t^3 + 2bt^2 + 4b^2 t/3 < (t+x)(t+y)(t+z) (t+a)^3, ]
and hence
Inequality () follows from this and ().
The proof of () uses Maclaurin's inequality and the inequality of arithmetic and geometric means to get
[
t+gg [(t+x)(t+y)(t+z)xyz]^1/3 = [(1+tx)(1+ty)
(1+tz)]^1/3 1 + th, ]
with equalities iff ( x=y=z ), whence
Two applications of () complete the proof of ().
Acknowledgment. We thank Arthur Gautesen for suggesting the use of uniform approximations.
