If the product of $n$ positive numbers is unity, then their sum is

If the product of $n$ positive numbers is unity, then their sum is

a positive integer
divisible by $n$
equal to $n + rac{1}{n}$
never less than $n$
This question was previously asked in
UPSC CAPF – 2017
The correct answer is D) never less than n.
Let the n positive numbers be $x_1, x_2, …, x_n$.
We are given that their product is unity: $x_1 \times x_2 \times \dots \times x_n = 1$.
We want to determine the property of their sum: $S = x_1 + x_2 + \dots + x_n$.
According to the Arithmetic Mean – Geometric Mean (AM-GM) inequality, for a set of non-negative numbers, the arithmetic mean is always greater than or equal to the geometric mean. Since the numbers are positive, this inequality applies:
$\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n}$
Substitute the given product into the inequality:
$\frac{S}{n} \ge \sqrt[n]{1}$
$\frac{S}{n} \ge 1$
$S \ge n$
The sum of the n positive numbers whose product is unity is always greater than or equal to n.
Equality holds if and only if all the numbers are equal. If $x_1 = x_2 = \dots = x_n = x$, and their product is 1, then $x^n = 1$. Since they are positive, $x$ must be 1. In this case, the sum is $n \times 1 = n$. If the numbers are not all equal, the sum is strictly greater than n. Therefore, the sum is never less than n.
The AM-GM inequality is a fundamental concept in mathematics often used to find minimum or maximum values or to prove inequalities. It states that for non-negative numbers $a_1, a_2, \dots, a_n$, $\frac{a_1 + a_2 + \dots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \dots a_n}$, with equality holding if and only if $a_1 = a_2 = \dots = a_n$.
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