If a a + 1 + b b + 1 + c c + 1 = 1 prove a b c ≤ 1 / 8Could I get some help with this?What I have tried-Making the denominators equal and adding to get another equation of the form N M = 1 then multiplying by M both the sides and simplifying to get 2 a b c + a b + b c + a c = 1 , from which we have, a b c &lt; 1 / 2But which obviously isn't enough to prove that a b c ≤ 1 / 8I also tried using Jensen's inequality which states that f ( E ( X ) ) ≥ E ( f ( X ) ) when f is a concave function and f ( E ( X ) ) ≤ E ( f ( X ) ) when f is a convex function, and where E ( X ) denotes the expectation of X ,but that didn't really help.

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If             a              a        +        1              +            b              b        +        1              +            c              c        +        1              =  1 prove   a  b  c  ≤  1      /    8Could I get some help with this?What I have tried-Making the denominators equal and adding to get another equation of the form             N      M        =  1 then multiplying by   M both the sides and simplifying to get   2  a  b  c  +  a  b  +  b  c  +  a  c  =  1 , from which we have,   a  b  c  &amp;lt;  1      /    2But which obviously isn&#039;t enough to prove that   a  b  c  ≤  1      /    8I also tried using Jensen&#039;s inequality which states that   f  (  E  (  X  )  )  ≥  E  (  f  (  X  )  ) when   f is a concave function and   f  (  E  (  X  )  )  ≤  E  (  f  (  X  )  ) when   f is a convex function, and where   E  (  X  ) denotes the expectation of X ,but that didn&#039;t really help.

Nkgopotsev1g · Accepted Answer

What you have (   2  a  b  c  +  a  b  +  b  c  +  c  a  =  1 and   a  ,  b  ,  c  &amp;gt;  0 ) does actually imply   a  b  c  ≤  1      /    8. This is because, by AM-GM,  a  b  +  b  c  +  c  a  ≥  3                    a        2                    b        2                    c        2              3  and so, writing   x  =  a  b  c, we have   1  ≥  2  x  +  3      x          2              /            3        . The RHS is clearly increasing in   x, and we have equality when   x  =  1      /    8, so any   x  &amp;gt;  1      /    8 violates this inequality.

Widersinnby7 · Accepted Answer

If you multiply both sides by   (  a  +  1  )  (  b  +  1  )  (  c  +  1  ) you&#039;ll obtain  3  a  b  c  +  2  a  b  +  2  a  c  +  2  b  c  +  a  +  b  +  c  =  (  a  +  1  )  (  b  +  1  )  (  c  +  1  )  ,which simplifies to  3  a  b  c  +  2  a  b  +  2  a  c  +  2  b  c  +  a  +  b  +  c  =  1  +  a  +  b  +  a  b  +  c  +  a  c  +  b  c  +  a  b  c  ,rearranging to  a  b  c  =            1      −      (      a      b      +      a      c      +      b      c      )        2    .Using AM-GM you get  a  b  +  a  c  +  b  c  ≥  3                    a        2                    b        2                    c        2              3    .Now set   u  =  a  b  c and you have  u  =            1      −                        u          2                3              2  which leads to  1  ≥  2  u  +  3      u          2              /            3        .Now you see that for   u  =  a  b  c  =  1      /    8 you have equality and for   u  =  a  b  c  &amp;lt;  1      /    8 you satisfy the inequality.

If a,b,c are positive real numbers and a/(a+1)+b/(b+1)+c/(c+1)=1 then prove that abc<=1/8 Could I get some help with this?

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