# Proving that the arithmetic and geometric means of a collection of non-negative numbers lies between their minimum and maximum values

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!

• $a_1\le y$, $a_2\le y,\ldots$ so $a_1a_2\cdots a_n\le y^n$. Does that help you show that the GM is $\le y$? – Angina Seng Jan 11 '19 at 5:19

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 $$k=1,2,3,\dots,n$$.

Just note that

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

GM: $$\sqrt[n]{a_1a_2\cdots a_n}\geq\sqrt[n]{x^n}= x$$

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

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