Solution:

Given:
a1 a2 a3.an=1

We need to find the sum of a1, a2, a3, and an.

Approach:
We can use the AM-GM inequality to solve this problem.

AM-GM Inequality:
The Arithmetic Mean - Geometric Mean (AM-GM) inequality states that the arithmetic mean of a set of non-negative numbers is always greater than or equal to the geometric mean of the same set of numbers.

For any set of non-negative numbers, a1, a2, ..., an, we have:
(a1 + a2 + ... + an)/n ≥ (a1a2...an)^(1/n)

Explanation:

Using the AM-GM inequality, we have:
(a1 + a2 + ... + an)/n ≥ (a1a2...an)^(1/n)

Multiplying both sides of the equation by n, we get:
a1 + a2 + ... + an ≥ n(a1a2...an)^(1/n)

Since a1a2...an = 1, we have:
a1 + a2 + ... + an ≥ n

Therefore, the sum of a1, a2, a3, and an is greater than or equal to n.

Answer:
The sum of a1, a2, a3, and an is greater than or equal to n, where n is the number of terms in the product a1a2...an, which is equal to 4 in this case.

Therefore, a1 + a2 + a3 + an ≥ 4.

Hence, the sum of a1, a2, a3, and an is greater than or equal to 4.