Inequality in 3 variables with a constraint condition












2












$begingroup$


To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers



Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which again after some algebra reduces to a²b²+b²c²+a²c² +2abc(a+b+c) > 2+3abc(a+b+c)=> a²b²+b²c²+a²c² > 2+abc . Now, this obviously true for positive a,b,c but I can't prove it rigorously this.






inequality contest-math cauchy-schwarz-inequality







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    2












    $begingroup$


    To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers



    Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which again after some algebra reduces to a²b²+b²c²+a²c² +2abc(a+b+c) > 2+3abc(a+b+c)=> a²b²+b²c²+a²c² > 2+abc . Now, this obviously true for positive a,b,c but I can't prove it rigorously this.






    inequality contest-math cauchy-schwarz-inequality







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers



      Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which again after some algebra reduces to a²b²+b²c²+a²c² +2abc(a+b+c) > 2+3abc(a+b+c)=> a²b²+b²c²+a²c² > 2+abc . Now, this obviously true for positive a,b,c but I can't prove it rigorously this.






      inequality contest-math cauchy-schwarz-inequality







      share|cite|improve this question











      $endgroup$




      To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers



      Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which again after some algebra reduces to a²b²+b²c²+a²c² +2abc(a+b+c) > 2+3abc(a+b+c)=> a²b²+b²c²+a²c² > 2+abc . Now, this obviously true for positive a,b,c but I can't prove it rigorously this.






      inequality contest-math cauchy-schwarz-inequality




      inequality contest-math cauchy-schwarz-inequality






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      edited Jan 9 at 22:13









      Michael Rozenberg

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      asked Jan 9 at 16:31









      Epsilon zeroEpsilon zero

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          $begingroup$

          If we arrange
          $$
          frac{ab+bc+ac}{abc} ge frac{2}{ab+bc+ac} +3
          $$           then we have
          $$
          (ab+bc+ac)^2 =sum_{text{cyc}}a^2b^2 +2abcge abc(2+3(ab+bc+ac))=2abc +3sum_{text{cyc}}a^2b^2c.
          $$           We need to show
          $$
          sum_{text{cyc}}a^2b^2ge 3sum_{text{cyc}}a^2b^2c.
          $$           This is equivalent to
          $$
          sum_{text{cyc}}a^2b^2(a+b+c)ge 3sum_{text{cyc}}a^2b^2c.
          $$           By rearrangement, we have
          $$
          sum_{text{cyc}}a^3b^2ge sum_{text{cyc}}a^2b^2c,
          $$          and
          $$
          sum_{text{cyc}}a^2b^3ge sum_{text{cyc}}a^2b^2c.
          $$           This proves the inequality.



          Another approach: We need to show
          $$
          sum_{text{cyc}}frac{a}{b}+sum_{text{cyc}}frac{a}{c}ge frac{2}{ab+bc+ca}.
          $$           By C-S, we have
          $$
          sum_{text{cyc}}frac{a}{b}sum_{text{cyc}}abge left(sum_{text{cyc}}aright)^2 =1,
          $$           and
          $$
          sum_{text{cyc}}frac{a}{c}sum_{text{cyc}}acge left(sum_{text{cyc}}aright)^2 =1.
          $$






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Because by C-S
            $$(ab+ac+bc)sum_{cyc}frac{a(b+c)}{bc}=sum_{cyc}acsum_{cyc}frac{a}{c}+sum_{cyc}absum_{cyc}frac{a}{b}geq$$
            $$geq(a+b+c)^2+(a+b+c)^2=2.$$






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            $endgroup$













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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

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              2












              $begingroup$

              If we arrange
              $$
              frac{ab+bc+ac}{abc} ge frac{2}{ab+bc+ac} +3
              $$               then we have
              $$
              (ab+bc+ac)^2 =sum_{text{cyc}}a^2b^2 +2abcge abc(2+3(ab+bc+ac))=2abc +3sum_{text{cyc}}a^2b^2c.
              $$               We need to show
              $$
              sum_{text{cyc}}a^2b^2ge 3sum_{text{cyc}}a^2b^2c.
              $$               This is equivalent to
              $$
              sum_{text{cyc}}a^2b^2(a+b+c)ge 3sum_{text{cyc}}a^2b^2c.
              $$               By rearrangement, we have
              $$
              sum_{text{cyc}}a^3b^2ge sum_{text{cyc}}a^2b^2c,
              $$              and
              $$
              sum_{text{cyc}}a^2b^3ge sum_{text{cyc}}a^2b^2c.
              $$               This proves the inequality.



              Another approach: We need to show
              $$
              sum_{text{cyc}}frac{a}{b}+sum_{text{cyc}}frac{a}{c}ge frac{2}{ab+bc+ca}.
              $$               By C-S, we have
              $$
              sum_{text{cyc}}frac{a}{b}sum_{text{cyc}}abge left(sum_{text{cyc}}aright)^2 =1,
              $$               and
              $$
              sum_{text{cyc}}frac{a}{c}sum_{text{cyc}}acge left(sum_{text{cyc}}aright)^2 =1.
              $$






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                If we arrange
                $$
                frac{ab+bc+ac}{abc} ge frac{2}{ab+bc+ac} +3
                $$                 then we have
                $$
                (ab+bc+ac)^2 =sum_{text{cyc}}a^2b^2 +2abcge abc(2+3(ab+bc+ac))=2abc +3sum_{text{cyc}}a^2b^2c.
                $$                 We need to show
                $$
                sum_{text{cyc}}a^2b^2ge 3sum_{text{cyc}}a^2b^2c.
                $$                 This is equivalent to
                $$
                sum_{text{cyc}}a^2b^2(a+b+c)ge 3sum_{text{cyc}}a^2b^2c.
                $$                 By rearrangement, we have
                $$
                sum_{text{cyc}}a^3b^2ge sum_{text{cyc}}a^2b^2c,
                $$                and
                $$
                sum_{text{cyc}}a^2b^3ge sum_{text{cyc}}a^2b^2c.
                $$                 This proves the inequality.



                Another approach: We need to show
                $$
                sum_{text{cyc}}frac{a}{b}+sum_{text{cyc}}frac{a}{c}ge frac{2}{ab+bc+ca}.
                $$                 By C-S, we have
                $$
                sum_{text{cyc}}frac{a}{b}sum_{text{cyc}}abge left(sum_{text{cyc}}aright)^2 =1,
                $$                 and
                $$
                sum_{text{cyc}}frac{a}{c}sum_{text{cyc}}acge left(sum_{text{cyc}}aright)^2 =1.
                $$






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  If we arrange
                  $$
                  frac{ab+bc+ac}{abc} ge frac{2}{ab+bc+ac} +3
                  $$                   then we have
                  $$
                  (ab+bc+ac)^2 =sum_{text{cyc}}a^2b^2 +2abcge abc(2+3(ab+bc+ac))=2abc +3sum_{text{cyc}}a^2b^2c.
                  $$                   We need to show
                  $$
                  sum_{text{cyc}}a^2b^2ge 3sum_{text{cyc}}a^2b^2c.
                  $$                   This is equivalent to
                  $$
                  sum_{text{cyc}}a^2b^2(a+b+c)ge 3sum_{text{cyc}}a^2b^2c.
                  $$                   By rearrangement, we have
                  $$
                  sum_{text{cyc}}a^3b^2ge sum_{text{cyc}}a^2b^2c,
                  $$                  and
                  $$
                  sum_{text{cyc}}a^2b^3ge sum_{text{cyc}}a^2b^2c.
                  $$                   This proves the inequality.



                  Another approach: We need to show
                  $$
                  sum_{text{cyc}}frac{a}{b}+sum_{text{cyc}}frac{a}{c}ge frac{2}{ab+bc+ca}.
                  $$                   By C-S, we have
                  $$
                  sum_{text{cyc}}frac{a}{b}sum_{text{cyc}}abge left(sum_{text{cyc}}aright)^2 =1,
                  $$                   and
                  $$
                  sum_{text{cyc}}frac{a}{c}sum_{text{cyc}}acge left(sum_{text{cyc}}aright)^2 =1.
                  $$






                  share|cite|improve this answer











                  $endgroup$



                  If we arrange
                  $$
                  frac{ab+bc+ac}{abc} ge frac{2}{ab+bc+ac} +3
                  $$                   then we have
                  $$
                  (ab+bc+ac)^2 =sum_{text{cyc}}a^2b^2 +2abcge abc(2+3(ab+bc+ac))=2abc +3sum_{text{cyc}}a^2b^2c.
                  $$                   We need to show
                  $$
                  sum_{text{cyc}}a^2b^2ge 3sum_{text{cyc}}a^2b^2c.
                  $$                   This is equivalent to
                  $$
                  sum_{text{cyc}}a^2b^2(a+b+c)ge 3sum_{text{cyc}}a^2b^2c.
                  $$                   By rearrangement, we have
                  $$
                  sum_{text{cyc}}a^3b^2ge sum_{text{cyc}}a^2b^2c,
                  $$                  and
                  $$
                  sum_{text{cyc}}a^2b^3ge sum_{text{cyc}}a^2b^2c.
                  $$                   This proves the inequality.



                  Another approach: We need to show
                  $$
                  sum_{text{cyc}}frac{a}{b}+sum_{text{cyc}}frac{a}{c}ge frac{2}{ab+bc+ca}.
                  $$                   By C-S, we have
                  $$
                  sum_{text{cyc}}frac{a}{b}sum_{text{cyc}}abge left(sum_{text{cyc}}aright)^2 =1,
                  $$                   and
                  $$
                  sum_{text{cyc}}frac{a}{c}sum_{text{cyc}}acge left(sum_{text{cyc}}aright)^2 =1.
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 9 at 17:21

























                  answered Jan 9 at 17:14









                  SongSong

                  11.4k628




                  11.4k628























                      0












                      $begingroup$

                      Because by C-S
                      $$(ab+ac+bc)sum_{cyc}frac{a(b+c)}{bc}=sum_{cyc}acsum_{cyc}frac{a}{c}+sum_{cyc}absum_{cyc}frac{a}{b}geq$$
                      $$geq(a+b+c)^2+(a+b+c)^2=2.$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Because by C-S
                        $$(ab+ac+bc)sum_{cyc}frac{a(b+c)}{bc}=sum_{cyc}acsum_{cyc}frac{a}{c}+sum_{cyc}absum_{cyc}frac{a}{b}geq$$
                        $$geq(a+b+c)^2+(a+b+c)^2=2.$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Because by C-S
                          $$(ab+ac+bc)sum_{cyc}frac{a(b+c)}{bc}=sum_{cyc}acsum_{cyc}frac{a}{c}+sum_{cyc}absum_{cyc}frac{a}{b}geq$$
                          $$geq(a+b+c)^2+(a+b+c)^2=2.$$






                          share|cite|improve this answer









                          $endgroup$



                          Because by C-S
                          $$(ab+ac+bc)sum_{cyc}frac{a(b+c)}{bc}=sum_{cyc}acsum_{cyc}frac{a}{c}+sum_{cyc}absum_{cyc}frac{a}{b}geq$$
                          $$geq(a+b+c)^2+(a+b+c)^2=2.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 9 at 22:13









                          Michael RozenbergMichael Rozenberg

                          101k1591193




                          101k1591193






























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