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Consider non-negative real numbers $a_1, a_2, a_3, ... , a_n$. How can I prove that both the arithmetic mean (AM) and the geometric mean (GM) of $a_1, a_2, a_3, ... , a_n$ are contained in the interval $[x, y]$, where $x = \text{minimum of} (a_1, a_2, a_3, ... , a_n)$ and $y = \text{maximum of} (a_1, a_2, a_3, ... , a_n)$?

I know that the GM AM inequality states that GM $\leq$ AM, so it would suffice to prove GM $\geq$ x and AM $\leq$ y. Am I correct so far, and if so, how should I proceed with the proof? Any hints or help in this direction would be greatly appreciated.

Thank you!

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  • $\begingroup$ $a_1\le y$, $a_2\le y,\ldots$ so $a_1a_2\cdots a_n\le y^n$. Does that help you show that the GM is $\le y$? $\endgroup$ – Lord Shark the Unknown Jan 11 at 5:19
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Yes, you are correct, it suffices to show that

i) $GM\geq x$, that is $$a_1 a_2 a_3 \cdots a_n\geq x\cdot x\cdot x\cdots x= x^n$$ which holds because $a_k\geq x=\min(a_1,a_2,a_3,\dots,a_n)\geq 0$ for $k=1,2,3,\dots,n$.

ii) $AM\leq y$, that is $$a_1+a_2+a_3 +\dots +a_n\leq y+ y+ y+\dots+ y=ny$$ which holds because $a_k\leq y=\max(a_1,a_2,a_3,\dots,a_n)$ for $k=1,2,3,\dots,n$.

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Just note that

  • $0\leq x \leq a_i \Rightarrow \sqrt[n]{x^n}\leq \sqrt[n]{a_1 \cdot \ldots \cdot a_n}$
  • $a_i \leq y \Rightarrow \frac{a_1 + \cdots + a_n}{n}\leq \frac{y + \cdots + y}{n} = y$
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GM: $\sqrt[n]{a_1a_2\cdots a_n}\geq\sqrt[n]{x^n}= x$

AM: $\frac{a_1+a_2+\cdots+ a_n}{n}\leq \frac{ny}{n}=y$

$y\geq(AM,GM)\geq x$

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