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Prove that $$ \left( \frac{a+b+c\cdots k}{n} \right) ^{a+b+c \cdots k} \lt a^a b^b c^c \cdots k^k $$

We can prove that $$ (1+x)^{1+x} (1-x)^{1-x} \gt 1$$ (if $x \lt 1$) and from there by assuming $x = \frac{u}{z}$ and then $u+z =a$ and $u-z =b$ we will have $$ \left( \frac{a+b}{2}\right)^{a+b} \lt a^a b^b$$ but how to extend this result upto $n$ number of terms?

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  • $\begingroup$ Have you tried taking logarithms on both sides ? $\endgroup$
    – Arnaud D.
    Commented Aug 26, 2020 at 13:09
  • $\begingroup$ @ArnaudD. Yes, it answers my question. Thanks. $\endgroup$ Commented Aug 26, 2020 at 13:57
  • $\begingroup$ Out of curiosity, did you find this in a book ? The author of the other questions mentions a book in the comments. I thought the formulation was a bit strange, so I edited the other question, but maybe I should have left it as it was if was copied from the book... $\endgroup$
    – Arnaud D.
    Commented Aug 26, 2020 at 14:00
  • $\begingroup$ @ArnaudD. Yes, it’s from Hall and Knight. $\endgroup$ Commented Aug 26, 2020 at 14:37
  • $\begingroup$ And I have used the same letters that are used in the book. $\endgroup$ Commented Aug 26, 2020 at 14:37

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