0
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E1 = 1/(2 k (c - m) m) (2 c dt m + c k m + c dt k m - k m^2 - 
      dt k m^2 - \[Sqrt](4 c^4 dt^2 + 8 c^4 dt^2 k + 4 c^4 dt^2 k^2 + 
         4 c^3 dt k m - 12 c^3 dt^2 k m + 4 c^4 dt^2 k m + 
         4 c^3 dt k^2 m - 12 c^3 dt^2 k^2 m + 4 c^4 dt^2 k^2 m - 
         4 c^2 dt k m^2 + 4 c^2 dt^2 k m^2 - 8 c^3 dt^2 k m^2 + 
         c^2 k^2 m^2 - 10 c^2 dt k^2 m^2 + 13 c^2 dt^2 k^2 m^2 - 
         12 c^3 dt^2 k^2 m^2 + 4 c^2 dt^2 k m^3 - 2 c k^2 m^3 + 
         8 c dt k^2 m^3 - 6 c dt^2 k^2 m^3 + 12 c^2 dt^2 k^2 m^3 + 
         k^2 m^4 - 2 dt k^2 m^4 + dt^2 k^2 m^4 - 
         4 c dt^2 k^2 m^4)) /. {dt -> 1};

E2 = 1/(2 k (c - m) m) (2 c dt m + c k m + c dt k m - k m^2 - 
      dt k m^2 + \[Sqrt](4 c^4 dt^2 + 8 c^4 dt^2 k + 4 c^4 dt^2 k^2 + 
         4 c^3 dt k m - 12 c^3 dt^2 k m + 4 c^4 dt^2 k m + 
         4 c^3 dt k^2 m - 12 c^3 dt^2 k^2 m + 4 c^4 dt^2 k^2 m - 
         4 c^2 dt k m^2 + 4 c^2 dt^2 k m^2 - 8 c^3 dt^2 k m^2 + 
         c^2 k^2 m^2 - 10 c^2 dt k^2 m^2 + 13 c^2 dt^2 k^2 m^2 - 
         12 c^3 dt^2 k^2 m^2 + 4 c^2 dt^2 k m^3 - 2 c k^2 m^3 + 
         8 c dt k^2 m^3 - 6 c dt^2 k^2 m^3 + 12 c^2 dt^2 k^2 m^3 + 
         k^2 m^4 - 2 dt k^2 m^4 + dt^2 k^2 m^4 - 
         4 c dt^2 k^2 m^4)) /. {dt -> 1};

Solve[Abs[E1] < 1 && Abs[E2] < 1, {c, m, k}]

I am not able to get reduced condition in simplest form of c, k, m.

I did try with 'Solve' but it is taking indefinite time. Help is solicited.

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  • $\begingroup$ It looks extremely complicated. Just try Reduce[E1 < 1, {c, m, k}] to see what I mean. $\endgroup$ – Lotus Feb 24 '17 at 7:24
  • $\begingroup$ Even Reducing one argument is taking too long time. Any smart idea is required. $\endgroup$ – Sk Sarif Hassan Feb 24 '17 at 7:46
  • $\begingroup$ What would help is if you have additional constraints for the variables. $\endgroup$ – Lotus Feb 24 '17 at 7:49
  • $\begingroup$ dt can be taken as 1 or 0.005. c, k, m are all real numbers. $\endgroup$ – Sk Sarif Hassan Feb 24 '17 at 8:02
  • $\begingroup$ It gets better if we have c > 0 etc..Reduce[{E1 < 1, c > 0, m > 0, k > 0}, {c, m, k}] $\endgroup$ – Lotus Feb 24 '17 at 8:04

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