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Given some constants $a_1,a_2,a_3,\cdots,a_n>0$ such that $a_1\cdot a_2\cdot a_3\cdot \cdots \cdot a_n=1$, prove by induction that $$\frac{a_1+a_2+\cdots+a_n}{n}\geq 1$$

I struggle to solve this task. I tried the following (I'll not elaborate the base cases, as they're trivial)

$$\frac{\sum^{n+1}_{i=1}a_i}{n+1}\geq1\Leftrightarrow\left(\frac{\sum^{n+1}_{i=1}a_i}{n+1}\right)^{n+1}\geq1^{n+1}=1\\\Leftrightarrow\left(\frac{1}{n}\sum^{n}_{i=1}a_i\right)^{n+1}\cdot\left(\frac{a_{n+1}+\sum^{n}_{i=1}a_i}{\frac{n+1}{n}\sum^{n}_{i=1}a_i}\right)^{n+1}\\ =\left(\frac{1}{n}\sum^{n}_{i=1}a_i\right)^{n+1}\cdot\left(\frac{a_{n+1}+n\frac{1}{n}\sum^{n}_{i=1}a_i+\frac{1}{n}\sum^n_{i=1}a_i-\frac{1}{n}\sum^n_{i=1}{a_i}}{\frac{n+1}{n}\sum^{n}_{i=1}a_i}\right)^{n+1}\\ =\left(\frac{1}{n}\sum^{n}_{i=1}a_i\right)^{n+1}\cdot\left(\frac{a_{n+1}+\frac{1}{n}(n+1)\sum^{n}_{i=1}a_i-\frac{1}{n}\sum^n_{i=1}{a_i}}{\frac{n+1}{n}\sum^{n}_{i=1}a_i}\right)^{n+1}\\ =\left(\frac{1}{n}\sum^{n}_{i=1}a_i\right)^{n+1}\cdot\left(1+\frac{a_{n+1}-\frac{1}{n}\sum^n_{i=1}{a_i}}{\frac{n+1}{n}\sum^{n}_{i=1}a_i}\right)^{n+1}\\ \geq\left(\frac{1}{n}\sum^{n}_{i=1}a_i\right)^{n+1}\cdot\left(1+(n+1)\frac{a_{n+1}-\frac{1}{n}\sum^n_{i=1}{a_i}}{\frac{n+1}{n}\sum^{n}_{i=1}a_i}\right)\\ =\left(\frac{1}{n}\sum^{n}_{i=1}a_i\right)^{n+1}\cdot\left(1+\frac{a_{n+1}}{\frac{1}{n}\sum^{n}_{i=1}a_i}-1\right)\\ =\left(\frac{1}{n}\sum^{n}_{i=1}a_i\right)^{n+1}\cdot a_{n+1}$$

.....ad this is where I am stuck now. How can I proceed from here? If I use the induction thesis then I'd end up with $\left(\frac{1}{n}\sum^{n}_{i=1}a_i\right)^{n+1}\cdot a_{n+1}\geq 1^n\cdot b_{n+1}$ which doesn't look promising at all. Maybe I made a mistake as well? How'd you prove it using the Bernoulli-Inequality?

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    $\begingroup$ The title and body of the question are slightly off. $\endgroup$ – Simply Beautiful Art Nov 13 '16 at 14:26
  • $\begingroup$ Seems hard to use induction, as $a_1\cdot a_2\cdots a_n=1$ and $a_1\cdot a_2\cdots a_{n+1}=1$ imply $a_{n+1}=1$. $\endgroup$ – Yves Daoust Nov 13 '16 at 14:33
  • $\begingroup$ @SimpleArt I fixed it $\endgroup$ – leonbloy Nov 13 '16 at 14:55
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Because of the AM-GM inequality you have

$$\dfrac{a_1+\cdots+a_n}{n}\ge \sqrt[n]{a_1\dots a_n}=\sqrt[n]{1}=1.$$ This, actually, is a proof that the inequality holds.

If you want to use induction you need to formulate properly the condition for $n.$ Note that, if $a_1\cdots a_n=x$ then $$\dfrac{a_1+\cdots+a_n}{n}\ge \sqrt[n]{a_1\dots a_n}=\sqrt[n]{x}.$$ (This is the inequality we need to use in the induction step. It is not necessary to use AM-GM inequality to get it. You only need to scale.) So, if $a_1\cdots a_na_{n+1}=1$ then

$$\dfrac{a_1+\cdots+a_n}{n}\ge \sqrt[n]{a_1\dots a_n}=\sqrt[n]{\dfrac{1}{a_{n+1}}}.$$ So,

$$\dfrac{a_1+\cdots+a_n+a_{n+1}}{n+1}=\dfrac{n}{n+1}\dfrac{a_1+\cdots+a_n}{n}+\dfrac{a_{n+1}}{n+1}\ge \dfrac{n}{n+1}\sqrt[n]{\dfrac{1}{a_{n+1}}}+\dfrac{a_{n+1}}{n+1}.$$ We need to show that $$ n\sqrt[n]{\dfrac{1}{x}}+x\ge n+1.$$ Write $x=t^n.$ We have to show that $$\dfrac{n}{t}+t^n\ge n+1.$$ But $f(t)=\dfrac{n}{t}+t^n$ has a minimum at $t=1.$ So, $f(t)\ge f(1)=n+1$ and we are done.

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  • $\begingroup$ This does not use Bernoulli $\endgroup$ – NaCl Nov 13 '16 at 14:28
  • $\begingroup$ I have editted the answer to use induction. $\endgroup$ – mfl Nov 13 '16 at 14:48
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I do not think that this can be done using Bernoulli. The problem is that Bernoulli's inequality is strict once the exponent is at least $2$, i.e. $(1+x)^n > 1 + nx$ for $n \geq 2$, hence applying Bernoulli would lead to a strict inequality. However, the inequality that you want to prove is not strict ($a_1 = \dots = a_n = 1$ leads to equality).

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  • $\begingroup$ Bernoulli inequality: $(1+x)^n\geq1+nx$ $\endgroup$ – NaCl Nov 13 '16 at 14:34
  • $\begingroup$ Yes, but once $n \geq 2$, we in fact have $(1+x)^n > 1 + nx$. $\endgroup$ – ToucanNapoleon Nov 13 '16 at 14:35
  • $\begingroup$ @HarrySmit Since $-1 \leq x$ for the Bernoulli inequality, it has to hold for $x = 0$. But this leads to equality for every $n$. $\endgroup$ – Nemo Nov 21 '16 at 5:34
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Let the $a_i$ in $a_1a_2…a_{n+1}=1$ be sorted from smallest to largest. Then $a_{n+1}\ge 1>0$. Consider $b_i=a_i·\sqrt[n]{a_{n+1}}$. Then $b_1b_2…b_n=1$ and by induction hypothesis $$ n\le b_1+b_2+…+b_n\iff a_1+a_2+…+a_n\ge n/\sqrt[n]{a_{n+1}} $$ Now one needs to show that $n/\sqrt[n]{a_{n+1}}\ge n+1-a_{n+1}$ or $$(a_{n+1})^{-\frac1n}\le 1-\frac1n(a_{n+1}-1)$$ to complete the induction step. By Bernoulli, $$ \left(1-\frac1n(a_{n+1}-1)\right)^n\ge 2-a_{n+1}=\frac{1-(1-a_{n+1})^2}{a_{n+1}} $$ from where this final inequality follows

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Proof by induction, letting aside the base case, because it is trivial.

We have $a_1,...,a_{n+1} > 0$ with $$ \prod_{j = 1}^{n+1} a_j = 1 \tag{1} $$ First, get your $a_j$'s into increasing order (by renaming them if necessary) $a_1 \leq a_2 \leq ... \leq a_{n+1}$ . Then it is also true that \begin{equation} \label{ineq} a_1 \leq 1 ~~~\text{and}~~~ 1 \leq a_{n+1} \tag{2} \end{equation}

Proof by contradiction Assume $ a_1 > 1 $ or $ 1 < a_{n+1} $ . Then, because of the ordering, it will also be true that $ \forall j ~~(a_j > 1$ or $ 1 < a_{j} ~)$ . But this implies $ \prod_{j=1}^{n+1} a_j \neq 1 $ , contradicting $(1) ~~~~\Box $

Thus, using the induction hypothesis on $(a_1 a_{n+1}) \cdot a_2 \cdot ... \cdot a_n = 1$ we get \begin{align} \Rightarrow ~~~~a_1 a_{n+1} + a_2 + &... + a_n \geq n \\ \Leftrightarrow ~~~~a_1 + ... + a_{n+1} &\geq n - a_1 a_{n+1} + a_1 + a_{n+1} \\ &= n + ( a_{n+1} - 1 ) ( 1 - a_1 ) + 1 \\ &\geq n + 1 \end{align} in the last step we used $(2)$, which tells us that $( a_{n+1} - 1 ) ( 1 - a_1 ) \geq 0$ . $~~~~~~~\Box$

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Following mfl's solution, the part where $f(t)$ is minimized at $t=1$ can be proven using Bernoulli's Inequality. Let $t=1+r$, so $r>-1$. Then, $$f(t)=\frac{n}{t}+t^n=n(1+r)^{-1}+(1+r)^n\geq n(1-r)+(1+nr)=n+1\,.$$ The equality holds iff $r=0$, or $t=1$.

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