Existence of a “scaling factor” in a inequality












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We're given the real numbers $a,b,c$ that satisfy the condition $ale b le c$.



Consider the case where $ble frac{a+b+c}{3}$, then we have $frac{a+b+c}{3} le frac{a+c}{2} le c$ and similarly $frac{a+b+c}{3} le frac{b+c}{2} le c$, Then there exist $lambda ,mu in [0,1]$ such that: $$frac{c+a}{2}=lambda c+(1-lambda)left(frac{a+b+c}{3}right)$$
and $$frac{b+c}{2}=mu c+(1-mu)left(frac{a+b+c}{3}right)$$
I can "intuitively" see why this should be true, but why do such $lambda$ and $mu$ must exist?






elementary-number-theory inequality







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    We're given the real numbers $a,b,c$ that satisfy the condition $ale b le c$.



    Consider the case where $ble frac{a+b+c}{3}$, then we have $frac{a+b+c}{3} le frac{a+c}{2} le c$ and similarly $frac{a+b+c}{3} le frac{b+c}{2} le c$, Then there exist $lambda ,mu in [0,1]$ such that: $$frac{c+a}{2}=lambda c+(1-lambda)left(frac{a+b+c}{3}right)$$
    and $$frac{b+c}{2}=mu c+(1-mu)left(frac{a+b+c}{3}right)$$
    I can "intuitively" see why this should be true, but why do such $lambda$ and $mu$ must exist?






    elementary-number-theory inequality







    share|cite|improve this question

























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      0







      We're given the real numbers $a,b,c$ that satisfy the condition $ale b le c$.



      Consider the case where $ble frac{a+b+c}{3}$, then we have $frac{a+b+c}{3} le frac{a+c}{2} le c$ and similarly $frac{a+b+c}{3} le frac{b+c}{2} le c$, Then there exist $lambda ,mu in [0,1]$ such that: $$frac{c+a}{2}=lambda c+(1-lambda)left(frac{a+b+c}{3}right)$$
      and $$frac{b+c}{2}=mu c+(1-mu)left(frac{a+b+c}{3}right)$$
      I can "intuitively" see why this should be true, but why do such $lambda$ and $mu$ must exist?






      elementary-number-theory inequality







      share|cite|improve this question













      We're given the real numbers $a,b,c$ that satisfy the condition $ale b le c$.



      Consider the case where $ble frac{a+b+c}{3}$, then we have $frac{a+b+c}{3} le frac{a+c}{2} le c$ and similarly $frac{a+b+c}{3} le frac{b+c}{2} le c$, Then there exist $lambda ,mu in [0,1]$ such that: $$frac{c+a}{2}=lambda c+(1-lambda)left(frac{a+b+c}{3}right)$$
      and $$frac{b+c}{2}=mu c+(1-mu)left(frac{a+b+c}{3}right)$$
      I can "intuitively" see why this should be true, but why do such $lambda$ and $mu$ must exist?






      elementary-number-theory inequality




      elementary-number-theory inequality






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      asked yesterday









      Spasoje Durovic

      34110




      34110






















          1 Answer
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          With $x = frac{a+b+c}{3}, y= frac{a+c}{2}, z= c$ your question becomes:




          If $x le y le z$ then there is a $lambda in [0, 1]$ such that
          $$ tag{*}
          y = lambda x + (1 - lambda) z , .
          $$




          If $x = z$ then any $lambda in [0, 1]$ will do, otherwise $(*)$ is equivalent
          to
          $$
          lambda = frac{z-y}{z-x} , ,
          $$
          which satisfies $0 le lambda le 1$.






          share|cite|improve this answer





















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            1 Answer
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            active

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            2














            With $x = frac{a+b+c}{3}, y= frac{a+c}{2}, z= c$ your question becomes:




            If $x le y le z$ then there is a $lambda in [0, 1]$ such that
            $$ tag{*}
            y = lambda x + (1 - lambda) z , .
            $$




            If $x = z$ then any $lambda in [0, 1]$ will do, otherwise $(*)$ is equivalent
            to
            $$
            lambda = frac{z-y}{z-x} , ,
            $$
            which satisfies $0 le lambda le 1$.






            share|cite|improve this answer


























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              With $x = frac{a+b+c}{3}, y= frac{a+c}{2}, z= c$ your question becomes:




              If $x le y le z$ then there is a $lambda in [0, 1]$ such that
              $$ tag{*}
              y = lambda x + (1 - lambda) z , .
              $$




              If $x = z$ then any $lambda in [0, 1]$ will do, otherwise $(*)$ is equivalent
              to
              $$
              lambda = frac{z-y}{z-x} , ,
              $$
              which satisfies $0 le lambda le 1$.






              share|cite|improve this answer
























                2












                2








                2






                With $x = frac{a+b+c}{3}, y= frac{a+c}{2}, z= c$ your question becomes:




                If $x le y le z$ then there is a $lambda in [0, 1]$ such that
                $$ tag{*}
                y = lambda x + (1 - lambda) z , .
                $$




                If $x = z$ then any $lambda in [0, 1]$ will do, otherwise $(*)$ is equivalent
                to
                $$
                lambda = frac{z-y}{z-x} , ,
                $$
                which satisfies $0 le lambda le 1$.






                share|cite|improve this answer












                With $x = frac{a+b+c}{3}, y= frac{a+c}{2}, z= c$ your question becomes:




                If $x le y le z$ then there is a $lambda in [0, 1]$ such that
                $$ tag{*}
                y = lambda x + (1 - lambda) z , .
                $$




                If $x = z$ then any $lambda in [0, 1]$ will do, otherwise $(*)$ is equivalent
                to
                $$
                lambda = frac{z-y}{z-x} , ,
                $$
                which satisfies $0 le lambda le 1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                Martin R

                27.2k33254




                27.2k33254






























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