Can the inequality a + b c + b + c a + c + a b ≥ 6 be proved with differentiation? with a,b,c&gt;0I could do it with letting x = a b , y = b c , z = c a , but I wonder if it is solvable somehow with differentiation.

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Can the inequality             a      +      b        c    +            b      +      c        a    +            c      +      a        b    ≥  6 be proved with differentiation? with a,b,c&amp;gt;0I could do it with letting   x  =      a    b  ,   y  =      b    c  ,   z  =      c    a  , but I wonder if it is solvable somehow with differentiation.

Cameron Benitez · Accepted Answer

Yes, you can do it by differentiation too. In your terms, the LHS becomes  x  +      1    x    +  y  +      1    y    +  z  +      1    z    .By differentiation, you can show that  t  +      1    t    ≥  2  ,    t  &amp;gt;  0.It gives the result.

Liam Keller · Accepted Answer

Consider you that you look for the minimum of function                    (1)                    Φ        =                              a            +            b                    c                +                              b            +            c                    a                +                              c            +            a                    b                    Compute the partial derivatives                    (2)                                          ∂            Φ                                d            a                          =        −                              b            +            c                                a            2                          +                  1          b                +                  1          c                                        (3)                                          ∂            Φ                                d            b                          =        −                              a            +            c                                b            2                          +                  1          a                +                  1          c                                        (4)                                          ∂            Φ                                d            c                          =        −                              a            +            b                                c            2                          +                  1          a                +                  1          b                    and say that all of them are equal to 0.From   (  2  ) solve the quadratic for   b. The two roots are      b    1    =            a      2        c      and        b    2    =  −  c    ⟹    b  =            a      2        c  since   a  ,  b  ,  c are positive.Plug   b  =            a      2        c   in   (  3  ) to get            (      a      +      c      )              (                  a          3                −                  c          3                )                            a        4            c        =  0    ⟹    c  =  a    ⟹    b  =  c  =  aPlug in (4) to get 0=0.Replace in (1) and get 6.

Can the inequality (a+b)/(c) + (b+c)/(a) + (c+a)/(b) >= 6 with a,b,c > 0 be proved with differentiation? I could do it with letting x=a/b, y=b/c, z=c/a, but I wonder if it is solvable somehow with differentiation.

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