Inequality. ${{sqrt{a}+sqrt{b}+sqrt{c}} over {2}} ge {{1} over {sqrt{a}}} + {{1} over {sqrt{b}}} + {{1} over...












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Question. If ${{a} over {1+a}}+{{b} over {1+b}}+{{c} over {1+c}}=2$ and $a$, $b$, $c$ are all positive real numbers, prove that



$${{sqrt{a}+sqrt{b}+sqrt{c}} over {2}} ge {{1} over {sqrt{a}}} + {{1} over {sqrt{b}}} + {{1} over {sqrt{c}}}$$



My approach. If we let



$$x:=frac{1}{1+a}, y:=frac{1}{1+b}, z:=frac{1}{1+c}$$



Then we can know that



$${{a} over {1+a}}+{{b} over {1+b}}+{{c} over {1+c}}=2 Leftrightarrow abc=a+b+c+2 Leftrightarrow x+y+z=1$$ (By definition of x, y, z)



And,



$$a=frac{1}{x}-1=frac{1-x}{x}=frac{y+z}{x} (because x+y+z=1)$$



$$therefore (a, b, c)=(frac{y+z}{x}, frac{z+x}{y}, frac{x+y}{z})$$



T.S. $$sum_{cyc}frac{sqrt{a}}{2}>sum_{cyc}frac{1}{sqrt{a}}$$



$$ Leftrightarrow sum_{cyc}(sqrt{a}-frac{2}{sqrt{a}}) ge 0$$



$$ Leftrightarrow sum_{cyc}(sqrt{frac{y+z}{x}}-2sqrt{frac{x}{y+z}}) ge 0$$



$$ Leftrightarrow sum_{cyc}frac{(y-x)+(z-x)}{sqrt{x(y+z)}}ge 0$$



$$ Leftrightarrow sum_{cyc}(x-y)(frac{1}{sqrt{y(z+x)}}-frac{1}{sqrt{x(y+z)}})ge 0$$



$$ Leftrightarrow sum_{cyc}(x-y)frac{sqrt{x(y+z)}-sqrt{y(z+x)}}{sqrt{xy(x+z)(y+z)}} ge 0$$



But I don't know the next stage.



What should I do?






inequality radicals substitution sos







share|cite|improve this question











$endgroup$

















    7












    $begingroup$


    Question. If ${{a} over {1+a}}+{{b} over {1+b}}+{{c} over {1+c}}=2$ and $a$, $b$, $c$ are all positive real numbers, prove that



    $${{sqrt{a}+sqrt{b}+sqrt{c}} over {2}} ge {{1} over {sqrt{a}}} + {{1} over {sqrt{b}}} + {{1} over {sqrt{c}}}$$



    My approach. If we let



    $$x:=frac{1}{1+a}, y:=frac{1}{1+b}, z:=frac{1}{1+c}$$



    Then we can know that



    $${{a} over {1+a}}+{{b} over {1+b}}+{{c} over {1+c}}=2 Leftrightarrow abc=a+b+c+2 Leftrightarrow x+y+z=1$$ (By definition of x, y, z)



    And,



    $$a=frac{1}{x}-1=frac{1-x}{x}=frac{y+z}{x} (because x+y+z=1)$$



    $$therefore (a, b, c)=(frac{y+z}{x}, frac{z+x}{y}, frac{x+y}{z})$$



    T.S. $$sum_{cyc}frac{sqrt{a}}{2}>sum_{cyc}frac{1}{sqrt{a}}$$



    $$ Leftrightarrow sum_{cyc}(sqrt{a}-frac{2}{sqrt{a}}) ge 0$$



    $$ Leftrightarrow sum_{cyc}(sqrt{frac{y+z}{x}}-2sqrt{frac{x}{y+z}}) ge 0$$



    $$ Leftrightarrow sum_{cyc}frac{(y-x)+(z-x)}{sqrt{x(y+z)}}ge 0$$



    $$ Leftrightarrow sum_{cyc}(x-y)(frac{1}{sqrt{y(z+x)}}-frac{1}{sqrt{x(y+z)}})ge 0$$



    $$ Leftrightarrow sum_{cyc}(x-y)frac{sqrt{x(y+z)}-sqrt{y(z+x)}}{sqrt{xy(x+z)(y+z)}} ge 0$$



    But I don't know the next stage.



    What should I do?






    inequality radicals substitution sos







    share|cite|improve this question











    $endgroup$















      7












      7








      7


      2



      $begingroup$


      Question. If ${{a} over {1+a}}+{{b} over {1+b}}+{{c} over {1+c}}=2$ and $a$, $b$, $c$ are all positive real numbers, prove that



      $${{sqrt{a}+sqrt{b}+sqrt{c}} over {2}} ge {{1} over {sqrt{a}}} + {{1} over {sqrt{b}}} + {{1} over {sqrt{c}}}$$



      My approach. If we let



      $$x:=frac{1}{1+a}, y:=frac{1}{1+b}, z:=frac{1}{1+c}$$



      Then we can know that



      $${{a} over {1+a}}+{{b} over {1+b}}+{{c} over {1+c}}=2 Leftrightarrow abc=a+b+c+2 Leftrightarrow x+y+z=1$$ (By definition of x, y, z)



      And,



      $$a=frac{1}{x}-1=frac{1-x}{x}=frac{y+z}{x} (because x+y+z=1)$$



      $$therefore (a, b, c)=(frac{y+z}{x}, frac{z+x}{y}, frac{x+y}{z})$$



      T.S. $$sum_{cyc}frac{sqrt{a}}{2}>sum_{cyc}frac{1}{sqrt{a}}$$



      $$ Leftrightarrow sum_{cyc}(sqrt{a}-frac{2}{sqrt{a}}) ge 0$$



      $$ Leftrightarrow sum_{cyc}(sqrt{frac{y+z}{x}}-2sqrt{frac{x}{y+z}}) ge 0$$



      $$ Leftrightarrow sum_{cyc}frac{(y-x)+(z-x)}{sqrt{x(y+z)}}ge 0$$



      $$ Leftrightarrow sum_{cyc}(x-y)(frac{1}{sqrt{y(z+x)}}-frac{1}{sqrt{x(y+z)}})ge 0$$



      $$ Leftrightarrow sum_{cyc}(x-y)frac{sqrt{x(y+z)}-sqrt{y(z+x)}}{sqrt{xy(x+z)(y+z)}} ge 0$$



      But I don't know the next stage.



      What should I do?






      inequality radicals substitution sos







      share|cite|improve this question











      $endgroup$




      Question. If ${{a} over {1+a}}+{{b} over {1+b}}+{{c} over {1+c}}=2$ and $a$, $b$, $c$ are all positive real numbers, prove that



      $${{sqrt{a}+sqrt{b}+sqrt{c}} over {2}} ge {{1} over {sqrt{a}}} + {{1} over {sqrt{b}}} + {{1} over {sqrt{c}}}$$



      My approach. If we let



      $$x:=frac{1}{1+a}, y:=frac{1}{1+b}, z:=frac{1}{1+c}$$



      Then we can know that



      $${{a} over {1+a}}+{{b} over {1+b}}+{{c} over {1+c}}=2 Leftrightarrow abc=a+b+c+2 Leftrightarrow x+y+z=1$$ (By definition of x, y, z)



      And,



      $$a=frac{1}{x}-1=frac{1-x}{x}=frac{y+z}{x} (because x+y+z=1)$$



      $$therefore (a, b, c)=(frac{y+z}{x}, frac{z+x}{y}, frac{x+y}{z})$$



      T.S. $$sum_{cyc}frac{sqrt{a}}{2}>sum_{cyc}frac{1}{sqrt{a}}$$



      $$ Leftrightarrow sum_{cyc}(sqrt{a}-frac{2}{sqrt{a}}) ge 0$$



      $$ Leftrightarrow sum_{cyc}(sqrt{frac{y+z}{x}}-2sqrt{frac{x}{y+z}}) ge 0$$



      $$ Leftrightarrow sum_{cyc}frac{(y-x)+(z-x)}{sqrt{x(y+z)}}ge 0$$



      $$ Leftrightarrow sum_{cyc}(x-y)(frac{1}{sqrt{y(z+x)}}-frac{1}{sqrt{x(y+z)}})ge 0$$



      $$ Leftrightarrow sum_{cyc}(x-y)frac{sqrt{x(y+z)}-sqrt{y(z+x)}}{sqrt{xy(x+z)(y+z)}} ge 0$$



      But I don't know the next stage.



      What should I do?






      inequality radicals substitution sos




      inequality radicals substitution sos






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 25 at 5:37









      Michael Rozenberg

      107k1895199




      107k1895199










      asked Jan 25 at 2:43









      coding1101coding1101

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          $begingroup$

          Now, use $$sqrt{x(y+z)}-sqrt{y(x+z)}=frac{x(y+z)-y(x+z)}{sqrt{x(y+z)}+sqrt{y(x+z)}}=frac{z(x-y)}{sqrt{x(y+z)}+sqrt{y(x+z)}}.$$






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            $begingroup$

            Now, use $$sqrt{x(y+z)}-sqrt{y(x+z)}=frac{x(y+z)-y(x+z)}{sqrt{x(y+z)}+sqrt{y(x+z)}}=frac{z(x-y)}{sqrt{x(y+z)}+sqrt{y(x+z)}}.$$






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              $begingroup$

              Now, use $$sqrt{x(y+z)}-sqrt{y(x+z)}=frac{x(y+z)-y(x+z)}{sqrt{x(y+z)}+sqrt{y(x+z)}}=frac{z(x-y)}{sqrt{x(y+z)}+sqrt{y(x+z)}}.$$






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                $begingroup$

                Now, use $$sqrt{x(y+z)}-sqrt{y(x+z)}=frac{x(y+z)-y(x+z)}{sqrt{x(y+z)}+sqrt{y(x+z)}}=frac{z(x-y)}{sqrt{x(y+z)}+sqrt{y(x+z)}}.$$






                share|cite|improve this answer









                $endgroup$



                Now, use $$sqrt{x(y+z)}-sqrt{y(x+z)}=frac{x(y+z)-y(x+z)}{sqrt{x(y+z)}+sqrt{y(x+z)}}=frac{z(x-y)}{sqrt{x(y+z)}+sqrt{y(x+z)}}.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 25 at 5:24









                Michael RozenbergMichael Rozenberg

                107k1895199




                107k1895199






























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