Let a , b , c &gt; 0 then show taht 9 a + b + c ≤ 2 ( 1 a + b + 1 b + c + 1 a + c ) ≤ 1 a + 1 b + 1 c .

Question

Let   a  ,  b  ,  c  &amp;gt;  0 then show taht       9          a      +      b      +      c        ≤  2      (          1              a        +        b              +          1              b        +        c              +          1              a        +        c                   )    ≤      1    a    +      1    b    +      1    c   .

Jerome Page · Accepted Answer

Step 1To prove       9          a      +      b      +      c        ≤  2      (          1              a        +        b              +          1              b        +        c              +          1              a        +        c              )   , we can observe that  a  +  b  +  c  =      1    2    (  (  a  +  b  )  +  (  a  +  c  )  +  (  b  +  c  )  )denote   x  =  (  a  +  b  )  ,  y  =  (  b  +  c  )  ,  z  =  (  a  +  c  ) and we get what we want to prove after simplify      9                  1        2            (      x      +      y      +      z      )        ≤  2  (      1    x    +      1    y    +      1    z    )Use Cauchy Schwarz Inequality directly as follows  (      1    x    +      1    y    +      1    z    )  (  x  +  y  +  z  )  ≥  (  1  +  1  +  1      )    2    =  9Orwe can use the GM-AM inequality only                     1        x            +              1        y            +              1        z              3    ≥  (      1          x      y      z            )                  1        3             and             x      +      y      +      z        3    ≥  (  x  y  z      )                  1        3             ,we can again get the above result.Orwe can use the GM-AM inequality                     1        x            +              1        y            +              1        z              3    ≥  (      1          x      y      z            )                  1        3             , then, just to prove      3          x      +      y      +      z        ≤  (      1          x      y      z            )                  1        3            which can be simplified as follow  (  x  +  y  +  z      )    3    ≥  27  x  y  z

Emanuel Keith · Accepted Answer

Step 1The inequality in question is just the majorization  (  a  ,  b  ,  c  )  ≻      (                  a        +        b            2        ,                  b        +        c            2        ,                  c        +        a            2        )    ≻      (                  a        +        b        +        c            3        ,                  a        +        b        +        c            3        ,                  a        +        b        +        c            3        )  applied to the convex function   t  ↦      1    t   .

Let a , b , c > 0 then show taht 9 a +

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