Invalid induction proof? Prove the following using mathematical induction. If $a_{1}, a_{1}, ... , a_{n}$ are positive real numbers such that if $$a_{1}a_{2}...a_{n} = 1 $$ then 
$$a_{1}+a_{2}+ ... + a_{n} \geq n$$ 
My proof: 
Basis case obvious. 
In the induction step I assume that $a_{1}a_{2}...a_{k} = 1$ and that $a_{1}+a_{2}+...+a_{k} \geq k$. Let $a_{k+1} $ be such that $a_{1}a_{12}...a_{k+1} = 1$. (this is not valid??) Then it follows that $a_{k+1} = 1$, and thus $$a_{1}+a_{1}+...+a_{k+1 \geq } k +a_{k+1} = k+1 $$  
 A: I think it's been established by other answers that there was a logical error in your proof. I hope you have rectified it so I'm not going to add to that. 
Here is my attempt. Not entirely sure of its validity though. 
Suppose $a_1 = 1$ for some $a_1 \in \Bbb R \implies a_1 \ge 1$ and hence the result is true for $n = 1$ 
Assume the result is true for every $k \lt n$. Then whenever $k \lt n,$
$$ a_1a_2...a_k = 1 \implies a_1 + a_2 + ... + a_k \ge k \;\; \text {for all $a_i \in \Bbb R$}$$  
Now we suppose $a_1a_2...a_n = 1$ and try to prove that $a_1 + a_2 + .. + a_n \ge n$
Suppose $a_1a_2..a_{n-1}a_n = 1 \implies a_1a_2..(a_{n-1}a_n) = 1$ which is actually a product of $n -1$ real numbers which is equal to $1$. Our induction hypothesis applies and we have that $$ a_1 + a_2 + ... + (a_{n-1}a_n) \ge n - 1 ---- (1)$$
Now as long as $n \gt 1$ we should be able to pick $a_{n-1}$ and $a_n$ (without loss of generality) such that $a_{n-1} \ge 1$ and $a_n \le 1$ because otherwise every $a_i \gt 1$ or every $a_i \lt 1$ leading to a contradiction. For these pairs, 
$$ (a_{n-1} - 1)( a_n - 1 ) \le 0  \implies a_{n-1} + a_n - 1 \ge a_{n-1}a_n $$
Combining this with inequality $(1)$ we have, $$ a_1 + a_2 + ... + a_{n - 1} + a_n - 1 \ge n - 1 \implies a_1 + a_2 + ... + a_{n - 1} + a_n \ge n $$
Q.E.D.
Bear in my mind I'm very much an amateur. Please let me know if there is an issue with my proof. Especially with the choice of $a_{n - 1}$ and $a_n$
A: The statement that you are proving with induction is
$$\forall n \in {\mathbb N} \forall a_1, \dots, a_n \in {\mathbb R}_+. a_1 a_2 \dots a_n = 1 \rightarrow a_1 + a_2 + \dots + a_n \geq n.$$
So the induction hypothesis is that $\forall a_1,\dots,a_n \in {\mathbb R}_+.a_1 a_2 \dots a_n = 1 \rightarrow a_1 + a_2 + \dots + a_n \geq n$ (i.e., quantified over all $a_i$, not just for particular $a_i$). You then take arbitrary $b_1, \dots, b_{n+1} \in {\mathbb R}_+$ with $b_1 b_2 \dots b_{n+1} = 1$ and try to deduce that $b_1 + b_2 + \dots + b_{n+1} \geq n+1$. In the process, you'll have to come up with some clever $a_1, \dots, a_n$ and use the induction hypothesis.
A: By AM-GM inequality for $n$ terms,  
$$\dfrac{a_1+a_2+a_3+\ldots+a_n}{n}\ge \sqrt[n]{a_1a_2\ldots a_n}$$
It is now given that $a_1a_2\ldots a_n=1$
So, we get, $a_1+a_2+\ldots+a_n\ge n$.
This completes our proof.
A: That's not the right way to use induction. Your argument only proved the case when every $a_k=1$.
Surly you can give a proof based on induction for this proposition, but it would be easier to apply the Inequality of arithmetic and geometric means directly.
A: By induction it can be proved like this:
Let $a_{1},\dots,a_{n+1}>0$ with $a_{1}\times\cdots\times a_{n}\times a_{n+1}=1$.
Then $\left(ca_{1}\right)\times\cdots\times\left(ca_{n}\right)=1$
for $c=a_{n+1}^{\frac{1}{n}}$ and by induction $\left(ca_{1}\right)+\cdots+\left(ca_{n}\right)\geq n$
leading to $a_{1}+\cdots+a_{n}+a_{n+1}\geq na_{n+1}^{-\frac{1}{n}}+a_{n+1}$.
So if we can prove that $nx+x^{-n}\geq n+1$ for each $x>0$ (here $x=a_{n+1}^{-\frac{1}{n}}$) then we are ready.
For $f\left(x\right)=nx+x^{-n}$
we find $f'\left(x\right)=n\left(1-x^{-n-1}\right)$. 
Then $f'\left(x\right)<0$
if $0<x<1$, $f'\left(1\right)=0$ and $f'\left(x\right)>0$ if $x>1$
showing that a minimum is achieved at $x=1$. 
This justifies the conclusion
that $nx+x^{-n}\geq f\left(1\right)=n+1$ for each $x>0$.
