I've been studying for my upcoming exams and I came across this exercise in the AM-GM inequality section:

If $a_n = \sqrt[n]{a}$ , $a \in \mathbb{R}$ , $ a > 0$ , $n \in \mathbb{N}$ then prove that:

$$ \dfrac{na}{1 + na} < a_n < 1 + \dfrac{a}{n}. $$

I've been trying to prove this for the past few hours but I am completely stuck at this point. Any help would be much appreciated!


A start: Consider the numbers $1,1,1,\dots,1$ ($n-1$ of them) and $a$. Their arithmetic mean is $\frac{n-1+a}{n}$ and their geometric mean is $\sqrt[n]{a}$. Now note that $1+\frac{a}{n}\gt \frac{n-1+a}{n}$.

  • $\begingroup$ Too tired. Thanks for the fast answer. $\endgroup$ – Veritas Dec 28 '14 at 18:05
  • $\begingroup$ You are welcome. It is only half an answer, I am leaving the other half to you. $\endgroup$ – André Nicolas Dec 28 '14 at 18:06
  • $\begingroup$ I solved it. The rest of the proofs in the book misdirected me and I completely missed the obvious approach. $\endgroup$ – Veritas Dec 28 '14 at 18:10
  • $\begingroup$ Good. The type of "trick" I used comes up medium often. $\endgroup$ – André Nicolas Dec 28 '14 at 18:13

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