# Extending the result $\left(\frac{a+b}{2}\right)^{a+b} \lt a^a b^b$ to $n$ number of terms. [duplicate]

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?

• Have you tried taking logarithms on both sides ? Aug 26, 2020 at 13:09
• @ArnaudD. Yes, it answers my question. Thanks. Aug 26, 2020 at 13:57
• 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... Aug 26, 2020 at 14:00
• @ArnaudD. Yes, it’s from Hall and Knight. Aug 26, 2020 at 14:37
• And I have used the same letters that are used in the book. Aug 26, 2020 at 14:37