Sum $S=a_1+a_2+...+a_{2015}$ Given positive real numbers $a_1,a_2,...,a_{2015}$ whose product  $\prod_{i=1}^{2015} a_i=1$. What can you say about their sum $S=a_1+a_2+...+a_{2015}$


*

*$S $ can be any positive number

*$1\leq S \leq 2015$

*$2015 \leq S$ and $S$ is unbounded above

*$2015 \leq S$ and $S$ is bounded above


I don't know how to attack such problems.  I think  $1 \leq S$ because the number of terms in the given series is odd so that $\forall a_i $ there is unique   $ a_j $ such that $a_i a_j=1$ so on pairing up one must be odd and it should be 1 but I am not sure about the rest
 A: From AM-GM inequality, we have
$$\dfrac{a_1+a_2 + \cdots a_{2015}}{2015} \geq \sqrt[2015]{a_1a_2\ldots a_{2015}} \implies a_1 + a_2 + \cdots + a_{2015} \geq 2015$$
There is no upper-bound on the sum, since taking $a_1 = M$, $a_2 = a_3 = \cdots = a_{2015} = \dfrac1{M^{1/2014}}$, we see that
$$a_1+a_2 + \cdots + a_{2015} > M$$
for any $M$.
A: BY AM $\geq GM$ inequality we can say that 
$$\frac{a_1+\cdots+a_{2015}}{2015}\geq (a_1a_2\cdots a_{2015})^{1/2015}=1$$
so that $S\geq 2015$.
Moreover, if each $a_i\in (0,1)$ then $S\leq 2015$ and in this way we can at least say that $S=2015$. 
If exactly one $a_i>1$ then there will be at least one $a_j$ such that $a_ia_j=1$ and all other $\prod\limits_{k\neq i,j}a_k=1$. Still we shall have $S\geq 2015$. 
For the upper bound we cannot find any positive real $\lambda$.
A: The third case is true...because by AM-GM inequality we have
$$
\frac S{2015}\ge 1
$$
Now choose numbers $a_1,a_2,\dotsc,a_{2014}$ as big as you want, then take
$$
a_{2015}=(a_1.a_2\dotsm a_{2014})^{-1}
$$
To prove that $S$ can be arbitrary big...
A: Take the case where $i=2$. Then $a_{1}a_{2}=1$ and we infer that $$a_{1}=\frac{1}{a_{2}}$$.
Thus
\begin{eqnarray}
a_{1}+a_{2} &=& a_{1}+\frac{1}{a_{1}} \\
            &=& \frac{a_{1}^{2}+1}{a_{1}} \\
\implies a_{1}+a_{2} &=& a_{2}(1+a_{1}^{2})
\end{eqnarray}
A: Given that this is a multiple-choice question, you don't actually need to invoke the AM-GM inequality.  Choices 2 and 4 can be ruled out by noting that $a_1=M$, $a_2=1/M$ and $a_i=1$ for $i\gt1$ gives $S\gt M$ no matter how large $M$ is.  And choice 1 is ruled out because $a_1a_2\cdots a_{2015}=1$ implies the $a_i$'s cannot all be less than $1$, so their sum $S$ is necessarily greater than $1$, hence cannot be any positive number.  That leaves choice 3 as the only possibility.  (If there'd been a choice 5, "none of the above," then you would need to invoke AM-GM.)
