If the product of n positive numbers is unity, then their sum is?
never less than n
The question asks about the sum of $n$ positive numbers given that their product is equal to unity (which means their product is 1). We are given $n$ positive numbers, let's call them $x_1, x_2, \dots, x_n$. We know that $x_i > 0$ for all $i=1, 2, \dots, n$, and their product is $x_1 \times x_2 \times \dots \times x_n = 1$. We need to find a property of their sum, $S = x_1 + x_2 + \dots + x_n$.
To relate the sum and product of positive numbers, a very useful tool is the Arithmetic Mean-Geometric Mean (AM-GM) inequality. This inequality states that for a set of $n$ non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean.
For $n$ positive numbers $x_1, x_2, \dots, x_n$, the AM is given by:
$$ \text{AM} = \frac{x_1 + x_2 + \dots + x_n}{n} $$
The GM is given by:
$$ \text{GM} = \sqrt[n]{x_1 x_2 \dots x_n} $$
The AM-GM inequality states:
$$ \text{AM} \ge \text{GM} $$
Substituting the expressions for AM and GM:
$$ \frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n} $$
Equality in the AM-GM inequality holds if and only if all the numbers are equal, i.e., $x_1 = x_2 = \dots = x_n$.
We are given that the product of the $n$ positive numbers is unity:
$$ x_1 x_2 \dots x_n = 1 $$
Substitute this into the right side (GM side) of the AM-GM inequality:
$$ \frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{1} $$
The $n$-th root of 1 is 1:
$$ \frac{x_1 + x_2 + \dots + x_n}{n} \ge 1 $$
Now, multiply both sides of the inequality by $n$ (since $n$ is the number of positive numbers, it is a positive integer, so multiplying by $n$ does not change the direction of the inequality):
$$ x_1 + x_2 + \dots + x_n \ge n $$
This result tells us that the sum of the $n$ positive numbers is greater than or equal to $n$.
The inequality $x_1 + x_2 + \dots + x_n \ge n$ means that the sum can be exactly equal to $n$ or it can be greater than $n$. It can never be less than $n$.
Therefore, the sum of $n$ positive numbers whose product is unity is always greater than or equal to $n$. In other words, it is never less than $n$.
Let's look at the given options in light of our finding that the sum $S \ge n$.
| Option | Analysis based on $Sum \ge n$ | Is it Always True? |
|---|---|---|
| a positive integer | Sum is always positive, but not always an integer (e.g., 2.5 for n=2) | No |
| divisible by n | Sum is not always divisible by n (e.g., 2.5 for n=2) | No |
| equal to $n + \frac{1}{n}$ | Sum is $\ge n$, not necessarily equal to $n + \frac{1}{n}$ (e.g., 10.1 for n=2) | No |
| never less than n | This is equivalent to $Sum \ge n$, which is proven by AM-GM inequality. | Yes |
| Concept | Description | Formula (for $x_1, \dots, x_n > 0$) | Condition for Equality |
|---|---|---|---|
| Arithmetic Mean (AM) | The average of a set of numbers. | $ \text{AM} = \frac{x_1 + \dots + x_n}{n} $ | N/A |
| Geometric Mean (GM) | The $n$-th root of the product of $n$ numbers. | $ \text{GM} = \sqrt[n]{x_1 \dots x_n} $ | N/A |
| AM-GM Inequality | For non-negative numbers, AM is greater than or equal to GM. | $ \frac{x_1 + \dots + x_n}{n} \ge \sqrt[n]{x_1 \dots x_n} $ | $x_1 = x_2 = \dots = x_n$ |
The AM-GM inequality is a fundamental tool in mathematics with many applications, especially in problems involving optimization (finding maximum or minimum values) and proving other inequalities. It highlights the relationship between additive and multiplicative properties of positive numbers.
For example, it can be used to show that for a fixed sum, the product of positive numbers is maximized when they are all equal. Conversely, for a fixed product, the sum of positive numbers is minimized when they are all equal, as demonstrated in this problem where the minimum sum is $n$ when the product is 1.
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