# Prove by induction $\frac{a_{1} + a_{2} + a_{3} +…+ a_{n}}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdot …a_n}$. [duplicate]

Let $a_1$ $a_2$,..., $a_n$ be positive numbers. Prove that $\frac{a_{1} + a_{2} + a_{3} +...+ a_{n}}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdot ....a_n}$. Mine is about trying to understand how can i use induction to make some justifications so this has nothing to do with solving inequalities

## marked as duplicate by Daniel W. Farlow, Winther, user147263, Joel Reyes Noche, N. F. TaussigMay 1 '15 at 1:10

• This is the GM-AM inequality. Look for the proof in the web. Surely it's in thousands of sites. – ajotatxe Apr 30 '15 at 22:52
• There are literally hundreds of proofs of this inequality. You wanted induction--my answer gives you induction. There are also numerous proofs of this inequality by induction. You really need to start searching before you post questions that have many duplicates. – Daniel W. Farlow Apr 30 '15 at 23:02
• the idea here is that i understand how this ineequality work ok i get that what i want it to know is that can i use induction to make my justification or should i used strong induction to prove this Ok.. I am not looking for people to solve my work for me when i post something i just want to see what other people are thinking about – user146269 Apr 30 '15 at 23:05
• ...and here is a big list of proofs (including by induction). Always try to look for a similar question before posting. This question has been answered at least 100 times on this site and it's not hard to find one of them. – Winther Apr 30 '15 at 23:07
• Ok i got that sorry for any problem that i cause – user146269 Apr 30 '15 at 23:07

Let $S(n)$ denote the statement $$S(n):\; \frac{x_1+x_2+\cdots+x_n}{n}\geq\sqrt[n]{x_1x_2\ldots x_n},\quad n\in\mathbb{N}.$$ Base step ($n=1$): The statement $S(1)$ says that $\frac{x_1}{1}\geq\sqrt[1]{x_1}$, which is true because $x_1 = x_1$.
Base step ($n=2$): The statement $S(2)$ says that $$\frac{x_1+x_2}{2}\geq\sqrt{x_1x_2},\tag{1}$$ which is true because $$a\leq x \leq b \longleftrightarrow a+b\geq x+\frac{ab}{x}, \qquad 0<a\leq b,\; x>0$$
Inductive step ($S(k)\to S(k+1)$): Fix some $k\geq 1$, where $k\in\mathbb{N}$. Assume that $$S(k):\; \frac{x_1+x_2+\cdots+x_k}{k} \geq \sqrt[k]{x_1x_2\ldots x_k}$$ holds. To be proved is that $$\frac{x_1+x_2+\cdots+x_{k+1}}{k+1}\geq\sqrt[k+1]{x_1x_2\ldots x_{k+1}}$$ follows. If $x_1 = x_2 = \cdots = x_{k+1}$, then the proof is done. If not, let $x_1x_2\ldots x_{k+1} = \rho^{k+1}$. Without loss of generality, assume that $x_1\leq x_i$ and $x_i \leq x_2$ for all $i$; that is, assume that $x_1 < \rho < x_2$. Beginning with the left side of $S(k+1)$ [excluding the $k+1$ divisor], \begin{align} x_1+x_2+\cdots+x_{k+1} &> \rho+\frac{x_1x_2}{\rho}+x_3+\cdots+x_k+x_{k+1}\tag{by $(1)$}\\ &\geq \rho+k\cdot\left(\sqrt[k]{\frac{x_1x_2}{\rho}x_3\ldots x_{k+1}}\right)\tag{by $S(k)$}\\ &= (k+1)\rho, \end{align} one arrives at the right side of $S(k+1)$ [with a $k+1$ multiple], thereby showing that $S(k+1)$ is also true, completing the inductive step. Thus, by mathematical induction, $S(n)$ is true for all $n\geq 1$, where $n\in\mathbb{N}$.