Proof of an inequality involving different exponentsI was asked the following interesting question but I can not come up with a proof.Let a , b , c be positive real numbers. Prove that ( ♣ ) a 3 b 3 + c 3 + b 3 a 3 + c 3 + c 3 a 3 + b 3 ≥ a 2 b 2 + c 2 + b 2 a 2 + c 2 + c 2 a 2 + b 2 One way to prove ( ♣ ) may be to consider the following function f ( x ) = a x b x + c x + b x a x + c x + c x a x + b x and prove that f ( x ) is a non-decreasing function on [ 2 , 3 ]. When we take its derivative, the formula looks quite tedious and we just can not continue on. So is there any better solution?

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Proof of an inequality involving different exponentsI was asked the following interesting question but I can not come up with a proof.Let   a  ,  b  ,  c be positive real numbers. Prove that                              (                      ♣                    )                                                  a            3                                              b              3                        +                          c              3                                      +                              b            3                                              a              3                        +                          c              3                                      +                              c            3                                              a              3                        +                          b              3                                      ≥                              a            2                                              b              2                        +                          c              2                                      +                              b            2                                              a              2                        +                          c              2                                      +                              c            2                                              a              2                        +                          b              2                                          One way to prove   (  ♣  ) may be to consider the following function  f  (  x  )  =            a      x                      b        x            +              c        x              +            b      x                      a        x            +              c        x              +            c      x                      a        x            +              b        x            and prove that   f  (  x  ) is a non-decreasing function on   [  2  ,  3  ]. When we take its derivative, the formula looks quite tedious and we just can not continue on. So is there any better solution?

Stevinivm · Accepted Answer

I think the solution with derivative is very smooth:      f    ′    (  x  )  =      ∑          c      y      c                          a        x            ln      ⁡      a      (              b        x            +              c        x            )      −              a        x            (              b        x            ln      ⁡      b      +              c        x            ln      ⁡      c      )              (              b        x            +              c        x                    )        2              =  =      ∑          c      y      c                          a        x            (              b        x            (      ln      ⁡      a      −      ln      ⁡      b      )      −              c        x            (      ln      ⁡      c      −      ln      ⁡      a      )      )              (              b        x            +              c        x                    )        2              =  =      ∑          c      y      c        (  ln  ⁡  a  −  ln  ⁡  b  )      (                            a          x                          b          x                            (                  b          x                +                  c          x                          )          2                      −                            a          x                          b          x                            (                  a          x                +                  c          x                          )          2                      )    =  =      ∑          c      y      c                          a        x                    b        x            (      ln      ⁡      a      −      ln      ⁡      b      )      (              a        x            −              b        x            )      (              a        x            +              b        x            +      2              c        x            )              (              a        x            +              c        x                    )        2            (              b        x            +              c        x                    )        2              ≥  0

Proof of an inequality involving different exponents I was asked the following interesting questio

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