# Inequality of arithmetic and geometric means [duplicate]

Prove that if $x_1,x_2,...,x_n$ are positive numbers, then $$\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+...\frac{1}{x_n}}\leq \sqrt[n]{x_1\cdot x_2\cdot ... \cdot x_n}\leq\frac{x_1+x_2+...+x_n}{n}$$

## marked as duplicate by Najib Idrissi, user147263, Martin R, Servaes, user149792Nov 29 '15 at 2:07

you can use the lemma: If $a_1, a_2, \cdots, a_n$ are real positive numbers such thet $a_1.a_2. \cdots . a_n=1$, then $$a_1+a_2+\cdots+a_n \geq n$$ occur the equality if, only if, $a_1=a_2=\cdots=a_n=1$. (You can proof this lemma by induction over $n$).

Now, lets proof the main result: If $a_1,a_2,\cdots,a_n$ are positive real numbers, then $$\sqrt[n]{a_1a_2\cdots a_n}\leq \frac{a_1+a_2+\cdots+a_n}{n}$$

Indeed, if $g=\sqrt[n]{a_1a_2\cdots a_n}$, follows that $$g^n=a_1a_2\cdots a_n \Rightarrow g.g.\cdots.g=a_1a_2\cdots a_n \Rightarrow \frac{a_1}{g}.\frac{a_2}{g}.\cdots.\frac{a_n}{g}=1$$ By lemma above, follows that $$\frac{a_1}{g}+\frac{a_2}{g}+\cdots+\frac{a_n}{g} \geq n \Rightarrow$$ $$\frac{a_1+a_2+\cdots+a_n}{n} \geq g \Rightarrow$$ $$\sqrt[n]{a_1a_2\cdots a_n}\leq \frac{a_1+a_2+\cdots+a_n}{n}$$ the equaly occur if, only if $$\frac{a_1}{g}=\frac{a_2}{g}=\cdots=\frac{a_n}{g}=1 \Leftrightarrow a_1=a_2=\cdots=a_n=g$$ i.e, the equality occur if, only if, every $a_i's$ are equals.

For proof that

$$\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}} \leq \sqrt[n]{x_1.\cdots. x_n}$$

we can use the inequality between the meas arthimetic and geometric with the numbers $\frac{1}{x_1},\cdots,\frac{1}{x_n}$. Indeed,

$$\sqrt[n]{\frac{1}{x_1}.\cdots.\frac{1}{x_n}}\leq \frac{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}{n}$$

therefore,

$$\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}} \leq \sqrt[n]{x_1.\cdots. x_n}$$