the side-coherence theorem
The side-coherence of the triple {a,b,c} is equivalent to either of the conditions
(1) Γ2 = (a+b+c)(a+b-c)(b+c-a)(c+a-b) > 0,
(2) Δ2 = 1 + 2cosh(a)cosh(b)cosh(c) - cosh2(a) - cosh2(b) - cosh2(c) > 0.
proof
The condition that the sum of any two of a,b,c is greater than the third can
be expressed in asymmetric form as |b-c| < a < (b+c).
(1) Since the quantities are non-negative, this is equivalent to
(b-c)2 < a2 < (b+c)2 i.e. a2 -(b-c)2 > 0 and (b+c)2 - a2 > 0.
Since (b+c)2 > (b-c)2, the expressions on the left cannot both be negative
Thus the condition is equivalent to ((b+c)2-a2)(a2-(b-c)2) > 0.
Expanding, we get Γ2 > 0.
(2) Since the quantities are non-negative, and cosh(x) is increasing on [0,∞)
and even, this is equivalent to cosh(b-c) < cosh(a) < cosh(b+c), and hence to
cosh(a) - cosh(b-c) > 0 and cosh(b+c)-cosh(a) > 0.
As cosh(b+c) > cosh(b-c), the expressions on the left cannot both be negative.
Thus the condition is equivalent to (cosh(a)-cosh(b-c))(cosh(b+c)-cosh(a)) > 0.
Using the addition formulae for cosh, and the fact that sinh2(x) = cosh2(x) - 1,
we eventually arrive at the condition Δ2 > 0.
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