In terms of a, b, c, the girths are {gab -> 4 a EllipticE[1 - b^2/a^2], gac -> 4 a EllipticE[1 - c^2/a^2], gbc -> 4 b EllipticE[1 - c^2/b^2]}

--at least with MAPLE9's definition of EllipticE, this is wrong [I divided by 4, but otherwise same as you]:

g := (a,b,c) -> (a*EllipticE(1 - b^2/a^2), a*EllipticE(1 - c^2/a^2), b*EllipticE(1 - c^2/b^2)); g(0.1, 5.0, 0.2); 0.000031428499 + 249.8998057 I, 0.02655964076 + 0.2498348128 I, 5.030078415

note nonreal "girths" I conjecture you really meant g := (a,b,c) -> (max(a,b)*EllipticE(1-min(a,b)^2/max(a,b)^2), max(a,c)*EllipticE(1-min(a,c)^2/max(a,c)^2), max(b,c)*EllipticE(1-min(b,c)^2/max(b,c)^2)); which at least returns positive real output for positive real input. Assuming you can correctly define g(a,b,c)... which should not be hard... then a solution to the posed problem would be got by running g on a lot of random numbers, producing a large set of triples, then using linear programming to find constants f1, f2, f3 such that f1*g1+f2*g2+f3*g3 always exceeded 0 for all triples (g1,g2,g3). Any such f1,f2,f3 would define an analogue of the triangle inequality as requested (at least in the limit of huge random data collection). We may wlog demand f2 and f3 have same sign while f1 has opposite sign, and by symmetry we may demand f2=f3. Just by hand considering these triples (1000,1000,0) and permutations (2*pi, 4, 4) and perms (1,1,1) I observe that g1<g2+g3 is obeyed but g1<0.9999*(g2+g3) can fail;; and g1>0*(g2+g3) is obeyed but g1>0.0001*(g2+g3) can fail. So I believe an inequality of the form g1<K*(g2+g3) is obeyed for some constant K>=1 and I do not know (but computer could pretty much tell us) what the minimum possible K is. The obvious guess is K=1. No inequality of the form g1>K*(g2+g3) is obeys for any K>0.

This is a routine problem for FindRoot, but it would be nice to see how a, b, c depend on the girths in the low eccentricity case, say, by reverting the series expansions of the girths. Julian, Corey, and I were surprised by the absence of multivariate reversion from Macsyma and Mathematica, and set about building our own.  After several hours of cruel hoodwinking by Mma's Series facility, we emerged having only solved the ellipsoid case. Surprisingly(?), it has at least eight (2^3) solutions, which may help to explain the shortage of multivariate reverters.  To fourth order,

{a -> (gab/(2 Pi))(1 - eac + eac^2 + ebc + (eac ebc)/2 - (3 ebc^2)/2 +      1/8 (-8 eac^3 - eac^2 ebc - 9 eac ebc^2 + 18 ebc^3) +      1/16 (16 eac^4 + 2 eac^3 ebc - 3 eac^2 ebc^2 + 42 eac ebc^3 - 57 ebc^4)),

 a -> (gac/(2 Pi))(1 + ebc + 3/2 (eac ebc - ebc^2) +      3/8 (eac^2 ebc - 7 eac ebc^2 + 6 ebc^3) -      3/16 (7 eac^2 ebc^2 - 26 eac ebc^3 + 19 ebc^4))}

{b -> (gab/(2 Pi))(1 + eac - (3 eac^2)/2 - ebc + (eac ebc)/2 + ebc^2 +      1/8 (18 eac^3 - 9 eac^2 ebc - eac ebc^2 - 8 ebc^3) +      1/16 (-57 eac^4 + 42 eac^3 ebc - 3 eac^2 ebc^2 + 2 eac ebc^3 +         16 ebc^4)),

 b -> (gac/(2 Pi))(1 + 2 eac - eac^2/2 - ebc - (eac ebc)/2 + ebc^2 +      1/8 (6 eac^3 - 5 eac^2 ebc + 7 eac ebc^2 - 8 ebc^3) +      1/16 (-21 eac^4 + 24 eac^3 ebc - 5 eac^2 ebc^2 - 14 eac ebc^3 +         16 ebc^4))}

{c -> (gab/(2 Pi))(1 - eac + eac^2 + eac^4 - ebc - (eac ebc)/2 - (eac^3 ebc)/      8 + ebc^2 - (5 eac^2 ebc^2)/16 - (eac ebc^3)/8 + ebc^4 +      1/8 (-8 eac^3 + eac^2 ebc + eac ebc^2 - 8 ebc^3)),

 c -> (gac/(2 Pi))(1 - ebc - (3 eac ebc)/2 - (3 eac^2 ebc)/8 +      ebc^2 + (9 eac ebc^2)/8 - (3 eac^2 ebc^2)/16 - ebc^3 - (      9 eac ebc^3)/8 + ebc^4)}

where {eac -> gab/gac - 1, ebc -> gab/gbc - 1}, which are small for low eccentricity.  E.g., for  {a -> 5., b -> 4., c -> 3.}, the girth formulæ give

{gab -> 28.3617, gac -> 25.527, gbc -> 22.1035}.

Plugging these back into the reversions: {{a -> 4.97828, a -> 4.98561},  {b -> 4.00689,  b -> 4.00371},  {c -> 3.0062,   c -> 3.00353}}. --rwg

-- Warren D. Smith http://RangeVoting.org  <-- add your endorsement (by clicking "endorse" as 1st step) and math.temple.edu/~wds/homepage/works.html