From: //noreply@axelvogt.de   
      
   On 06.03.2016 08:13, clicliclic@freenet.de wrote:   
   >   
   > Let a,b,c,h,k be real numbers. Can the following be expressed more   
   > compactly, perhaps by means of Cylindrical Algebraic Decomposition?   
   >   
   > h>=0 AND k>=0 AND ((2*a*b*c+k*(b^2-a^2)+b*(a^2+b^2)>=0 AND b<=0)   
   >   OR ((2*a*b*c+k*(b^2-a^2)+b*(a^2+b^2)<=0 AND a>=0)   
   >   AND (2*b*c*(a-h)-h^2*(k+b)+2*a*h*k+k*(b^2-a^2)+b*(a^2+b^2)<=0   
   >   AND a-h<=0) AND b+k>=0) OR (2*c*a*b+k*(b^2-a^2)+b*(a^2+b^2)<=0   
   >   AND 2*b*c*(a-h)-h^2*(k+b)+2*a*h*k+k*(b^2-a^2)+b*(a^2+b^2)>=0   
   >   AND b>=0))   
      
   I would write (0 <= h and 0 <= k and (T1 or T2 or T3)) where   
      
   T1 := 0 <= 2*a*b*c+k*(-a^2+b^2)+b*(a^2+b^2) and b <= 0   
   T2 := 2*a*b*c+k*(-a^2+b^2)+b*(a^2+b^2) <= 0 and 0 <= a and   
          2*b*c*(a-h)-h^2*(k+b)+2*a*h*k+k*(-a^2+b^2)+b*(a^2+b^2) <= 0 and   
          a-h <= 0 and 0 <= k+b   
   T3 := 2*a*b*c+k*(-a^2+b^2)+b*(a^2+b^2) <= 0 and   
          0 <= 2*b*c*(a-h)-h^2*(k+b)+2*a*h*k+k*(-a^2+b^2)+b*(a^2+b^2) and   
          0 <= b   
      
   and now abbreviate by r = 2*a*b*c+k*(b^2-a^2)+b*(a^2+b^2) to shorten it.   
      
   And perhaps re-arrange according to 0 <= a resp 0 <= b.   
      
   But may be you want something different.   
      
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