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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. Taussig May 1 '15 at 1:10

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    $\begingroup$ This is the GM-AM inequality. Look for the proof in the web. Surely it's in thousands of sites. $\endgroup$ – ajotatxe Apr 30 '15 at 22:52
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    $\begingroup$ 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. $\endgroup$ – Daniel W. Farlow Apr 30 '15 at 23:02
  • $\begingroup$ 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 $\endgroup$ – user146269 Apr 30 '15 at 23:05
  • $\begingroup$ ...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. $\endgroup$ – Winther Apr 30 '15 at 23:07
  • $\begingroup$ Ok i got that sorry for any problem that i cause $\endgroup$ – 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 $n\in\mathbb{N}$.


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