# prove of sum of n real numbers greater than n given product is one [duplicate]

If the product of $n$ positive real numbers is $1$.Then prove that their sum is never less than $n$.
Let $A$ be the set of our $n$ numbers. If the product of $n$ real numbers is $1$, there must be a number $n_k\in A$ such that $n_k\ge\frac {1}{k}$ for every $k\le 1$, $k\in A$. Hence $$\sum_{i=0}^\frac{n}{2}n_k+k\ge \sum_{i=2j}^n 2=n$$ where $j=0,1,2,3...$. The only thing you need to think about now is where the $2$ in the sum is coming from, you should be able to do so.