The inequality you're referring to is:

$(1 + a_1)(1 + a_2)\dots(1 + a_n) \geq 2^n$

This inequality appears to be a form of AM-GM inequality (Arithmetic Mean - Geometric Mean inequality). Let's analyze it under the assumption that $a_1, a_2, \dots, a_n$ are non-negative real numbers.

Applying the AM-GM Inequality:

The AM-GM inequality states that for non-negative numbers $x_1, x_2, \dots, x_n$,

$\frac{x_1 + x_2 + \dots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \dots x_n}$

In our case, let's take $x_i = 1 + a_i$ where $a_i \geq 0$. Then we apply AM-GM:

$\frac{(1 + a_1) + (1 + a_2) + \dots + (1 + a_n)}{n} \geq \sqrt[n]{(1 + a_1)(1 + a_2) \dots (1 + a_n)}$

Simplifying the left-hand side:

$\frac{n + a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{(1 + a_1)(1 + a_2) \dots (1 + a_n)}$

This simplifies to:

$1 + \frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{(1 + a_1)(1 + a_2) \dots (1 + a_n)}$

By the AM-GM inequality, this means:

$(1 + a_1)(1 + a_2)\dots(1 + a_n) \geq 2^n$

The inequality holds with equality if and only if $a_1 = a_2 = \dots = a_n = 1$.

Thus, for general non-negative $a_i$, the inequality will always hold true, and if all $a_i = 1$, the inequality becomes equality.

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Would you like further details or clarification on this result?

Here are 5 related questions to deepen your understanding:

1. What conditions lead to equality in the AM-GM inequality?
2. How does the AM-GM inequality apply to other algebraic problems?
3. Can you generalize this result to cases where $a_i$ are negative?
4. How does this inequality relate to other inequalities like Cauchy-Schwarz?
5. What happens if we increase the number of terms $n$ in this inequality?

Tip: The AM-GM inequality is a powerful tool for proving various inequalities and can often help simplify complex expressions into manageable forms.