# Help with simplification

   (1/(1 + a + c) + 1/(1 + b + c) - 3/(3 + 2*a + 2*b + 2*c))*
(1 - c - ((a*b - c)*(a^2 + b^2 + b*(-1 + c) + (-1 + c)*c +
a*(-1 + b + c - 3*b*c)))/(-6*a*b*c + c^2 +
b^2*(1 + c^2) +
a^2*(1 + b^2 + c^2))) + (1/(1 + a + b) + 1/(1 + b + c) -
3/(3 + 2*a + 2*b + 2*c))*
(1 - b + ((b - a*c)*(a^2 + b^2 + b*(-1 + c) + (-1 + c)*c +
a*(-1 + b + c - 3*b*c)))/(-6*a*b*c + c^2 +
b^2*(1 + c^2) +
a^2*(1 + b^2 + c^2))) + (1/(1 + a + b) + 1/(1 + a + c) -
3/(3 + 2*a + 2*b + 2*c))*
(1 - a + ((a - b*c)*(a^2 + b^2 + b*(-1 + c) + (-1 + c)*c +
a*(-1 + b + c - 3*b*c)))/(-6*a*b*c + c^2 +
b^2*(1 + c^2) +
a^2*(1 + b^2 + c^2)))


with

1 - a^2 - b^2 + 2 a b c - c^2=0


Thanks a lot.

• When I scrape-n-paste your code I find pale red ) and ( showing mismatches. Can you possible scrape-n-paste the code from the screen into your notebook and correct those? Thanks
– Bill
Mar 22 '20 at 5:50
• @Bill Okay, revised. Thanks. Mar 22 '20 at 6:09
• How do you know there is much simpler expression when 1 - a^2 - b^2 + 2 a b c - c^2 == 0? when you do Simplify[expr,1 - a^2 - b^2 + 2 a b c - c^2 == 0] it does not simplify. Mar 22 '20 at 6:28
• Or could anyone simplify the expression to an expression whose sign can be determined for all a, b, c satisfying $1 - a^2 - b^2 + 2a b c - c^2=0$, $a+b>-1$, $a+c>-1$, and $b<0$? Thanks. Mar 22 '20 at 6:29
• @Bill Thank you for your comment, but could you transform it to an expression whose sign can be determined for all a, b, c satisfying $1 - a^2 - b^2 + 2a b c - c^2=0$, $a+b>-1$, $a+c>-1$, and $b<0$? Thanks a lot.. Mar 22 '20 at 6:36

Edit (now regarding all conditions)

Seems impossible to find a simple form of the given expression due to the complicated form.

Regard the hypersurface where expr is greater or lower zero with regard of all conditions.

cpg = ContourPlot3D[1 - a^2 - b^2 + 2 a b c - c^2 == 0,
{a, -1, 10}, {b, -10, 0}, {c, -10, 10}, PlotPoints -> 100, ImageSize -> 300,
RegionFunction -> Function[{a, b, c},
a + b > -1 && a + c > -1 && b < 0 && expr[a, b, c] > 0]]


cpl = ContourPlot3D[1 - a^2 - b^2 + 2 a b c - c^2 == 0,
{a, -1, 10}, {b, -10, 0}, {c, -10, 10}, PlotPoints -> 100, ImageSize -> 300,
RegionFunction -> Function[{a, b, c},
a + b > -1 && a + c > -1 && b < 0 && expr[a, b, c] < 0]]


• Thank you for your plot, and from your plot, can you see that the values of the expression is always >0 for all a, b and c satisfying a + b > -1 && a + c > -1 && b < 0? Thanks. Mar 22 '20 at 9:33
• and so, can you show that the the expression is always >0 for all a, b and c satisfying a + b > -1 && a + c > -1 && b < 0 and 1 - a^2 - b^2 + 2 a b c - c^2=0 by transforming the expression? Thanks a lot. Mar 22 '20 at 9:35
• and so, can you show that the the expression is always >0 for all a, b and c satisfying a + b > 0&& a + c > 0 && b < 0 and 1 - a^2 - b^2 + 2 a b c - c^2=0 by transforming the expression? Thanks a lot. Mar 22 '20 at 10:15
• This {{amin, amax}, {bmin, bmax}, {cimn, cmax}} = {First@ Minimize[{#, a + b > -1 && a + c > -1 && b < 0}, {a, b, c}], First@Maximize[{#, a + b > -1 && a + c > -1 && b < 0}, {a, b, c}, Reals]} & /@ {a, b, c} // Quiet  shows the minima/maxima for a,b,c. Yileds {{-1, \[Infinity]}, {-\[Infinity], 0}, {-\[Infinity], \[Infinity]}}  . Mar 22 '20 at 11:30
• If you change to a + b > 0 && a + c > 0 && b < 0  , you get {{0, \[Infinity]}, {-\[Infinity], 0}, {-\[Infinity], \[Infinity]}}  . Mar 22 '20 at 11:34