Prove that $a+b+cle 0$

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$a,b,c$ are integer numbers different from zero and
$$fracab+c^2=fraca+c^2b$$
Prove that $a+b+cle 0$.
I know how to prove that $a+b<0$ but do not know what about $c$.










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    up vote
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    favorite












    $a,b,c$ are integer numbers different from zero and
    $$fracab+c^2=fraca+c^2b$$
    Prove that $a+b+cle 0$.
    I know how to prove that $a+b<0$ but do not know what about $c$.










    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      $a,b,c$ are integer numbers different from zero and
      $$fracab+c^2=fraca+c^2b$$
      Prove that $a+b+cle 0$.
      I know how to prove that $a+b<0$ but do not know what about $c$.










      share|cite|improve this question















      $a,b,c$ are integer numbers different from zero and
      $$fracab+c^2=fraca+c^2b$$
      Prove that $a+b+cle 0$.
      I know how to prove that $a+b<0$ but do not know what about $c$.







      linear-algebra algebra-precalculus inequality






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      edited Sep 1 at 9:19









      greedoid

      28k93776




      28k93776










      asked Sep 1 at 6:54









      math.trouble

      335




      335




















          3 Answers
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          Since $c in mathbbZ$, we have $c le |c|le c^2$



          $$0=(a+b)c^2+c^4$$



          $$a+b+c^2=0$$



          $$a+b+c le a+b+c^2=0$$






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            up vote
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            We have $$ab=(a+c^2)(b+c^2)\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2le -c$$or equivalently $$a+b+cle 0$$






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              From given equation we get $a+b+c^2 =0$ (since $cne 0$), so we have $$a+b+c= c-c^2 = c(1-c) leq 0$$



              If $c>0$ then $1-cleq 0$ so $c(1-c)leq 0$ and



              if $c<0$ then $1-c>0$ so $c(1-c)<0$






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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                4
                down vote



                accepted










                Since $c in mathbbZ$, we have $c le |c|le c^2$



                $$0=(a+b)c^2+c^4$$



                $$a+b+c^2=0$$



                $$a+b+c le a+b+c^2=0$$






                share|cite|improve this answer
























                  up vote
                  4
                  down vote



                  accepted










                  Since $c in mathbbZ$, we have $c le |c|le c^2$



                  $$0=(a+b)c^2+c^4$$



                  $$a+b+c^2=0$$



                  $$a+b+c le a+b+c^2=0$$






                  share|cite|improve this answer






















                    up vote
                    4
                    down vote



                    accepted







                    up vote
                    4
                    down vote



                    accepted






                    Since $c in mathbbZ$, we have $c le |c|le c^2$



                    $$0=(a+b)c^2+c^4$$



                    $$a+b+c^2=0$$



                    $$a+b+c le a+b+c^2=0$$






                    share|cite|improve this answer












                    Since $c in mathbbZ$, we have $c le |c|le c^2$



                    $$0=(a+b)c^2+c^4$$



                    $$a+b+c^2=0$$



                    $$a+b+c le a+b+c^2=0$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 1 at 7:01









                    Siong Thye Goh

                    81.8k1456104




                    81.8k1456104




















                        up vote
                        2
                        down vote













                        We have $$ab=(a+c^2)(b+c^2)\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2le -c$$or equivalently $$a+b+cle 0$$






                        share|cite|improve this answer
























                          up vote
                          2
                          down vote













                          We have $$ab=(a+c^2)(b+c^2)\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2le -c$$or equivalently $$a+b+cle 0$$






                          share|cite|improve this answer






















                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            We have $$ab=(a+c^2)(b+c^2)\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2le -c$$or equivalently $$a+b+cle 0$$






                            share|cite|improve this answer












                            We have $$ab=(a+c^2)(b+c^2)\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2le -c$$or equivalently $$a+b+cle 0$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Sep 1 at 7:00









                            Mostafa Ayaz

                            10.3k3730




                            10.3k3730




















                                up vote
                                2
                                down vote













                                From given equation we get $a+b+c^2 =0$ (since $cne 0$), so we have $$a+b+c= c-c^2 = c(1-c) leq 0$$



                                If $c>0$ then $1-cleq 0$ so $c(1-c)leq 0$ and



                                if $c<0$ then $1-c>0$ so $c(1-c)<0$






                                share|cite|improve this answer
























                                  up vote
                                  2
                                  down vote













                                  From given equation we get $a+b+c^2 =0$ (since $cne 0$), so we have $$a+b+c= c-c^2 = c(1-c) leq 0$$



                                  If $c>0$ then $1-cleq 0$ so $c(1-c)leq 0$ and



                                  if $c<0$ then $1-c>0$ so $c(1-c)<0$






                                  share|cite|improve this answer






















                                    up vote
                                    2
                                    down vote










                                    up vote
                                    2
                                    down vote









                                    From given equation we get $a+b+c^2 =0$ (since $cne 0$), so we have $$a+b+c= c-c^2 = c(1-c) leq 0$$



                                    If $c>0$ then $1-cleq 0$ so $c(1-c)leq 0$ and



                                    if $c<0$ then $1-c>0$ so $c(1-c)<0$






                                    share|cite|improve this answer












                                    From given equation we get $a+b+c^2 =0$ (since $cne 0$), so we have $$a+b+c= c-c^2 = c(1-c) leq 0$$



                                    If $c>0$ then $1-cleq 0$ so $c(1-c)leq 0$ and



                                    if $c<0$ then $1-c>0$ so $c(1-c)<0$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Sep 1 at 7:34









                                    greedoid

                                    28k93776




                                    28k93776



























                                         

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