From: danielpf@gmail.com   
      
   Le lundi 21 août 2023 à 23:06:21 UTC+2, DP a écrit :   
   > Le dimanche 20 août 2023 à 18:39:42 UTC+2, nob...@nowhere.invalid a écrit   
   :    
   > > Is there a simple way to show algebraically that:    
   > >    
   > > b^2 >= (a - c)^2 AND a^2 >= b^2    
   > >    
   > > implies:    
   > >    
   > > c^2 >= (a + b)^2 OR c^2 >= (a - b)^2    
   > >    
   > > where ">=" denotes "greater or equal"?    
   > >    
   > > Martin.   
   > Taking a = 2, b = 1, and c = -2 gives a counterexample as the first   
   proposition is false, while the second true.    
   >    
   > Dan   
      
   Sorry I misunderstood the problem, which is if the first proposition is true   
   then the second is true as well.   
      
   To show that let x = a+b, and y = a-b.  Then from the first inequation it   
   comes after simplification   
      (c-x)(c-y) <= 0.   
      
   So there are two cases:   
   1) c <= x AND c >= y,   
   2) c >= x AND c <= y.   
      
   So either 1) is true and in this case   
      c >= y, that is c >=  a-b, c^2 >= (a-b)^2,   
   or 2) is true and   
      c >= a+b, c^2 >= (a+b)^2.   
      
   Note that one could more completely say that   
   either  c^2 >= (a-b)^2 AND c^2 <= (a+b)^2   OR  c^2 >= (a+b)^2 AND c^2 <=   
   (a-b)^2.   
      
   One could also keep the original inequations 1) and 2),   
     c >=  a-b AND c <= a+b   OR  c >= a+b AND c <= a-b.   
      
   Also it seems I didn't use the second inequation  a^2 >= b^2   
   which yields  xy >= 0 (x and y have the same sign).   
      
      Dan   
      
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