Maximizing Product abcd with Constraint

The question asks for the maximum value of the product $abcd$ given the condition $a^2 + b^2 + c^2 + d^2 = 1$. We need to find the largest possible value for $abcd$.

Applying AM-GM Inequality

We can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. For non-negative numbers $x_1, x_2, ..., x_n$, the inequality is:

$ \frac{x_1 + x_2 + ... + x_n}{n} \ge \sqrt[n]{x_1 x_2 ... x_n} $

Let's apply this to the terms $a^2, b^2, c^2, d^2$. These terms are non-negative.

$ \frac{a^2 + b^2 + c^2 + d^2}{4} \ge \sqrt[4]{a^2 b^2 c^2 d^2} $

Using the Given Constraint

We are given that $a^2 + b^2 + c^2 + d^2 = 1$. Substituting this into the inequality:

$ \frac{1}{4} \ge \sqrt[4]{(abcd)^2} $

The term $\sqrt[4]{(abcd)^2}$ simplifies to $\sqrt{|abcd|}$. So the inequality becomes:

$ \frac{1}{4} \ge \sqrt{|abcd|} $

Deriving the Maximum Product Value

To find the maximum value of $|abcd|$, we square both sides of the inequality:

$ \left(\frac{1}{4}\right)^2 \ge (\sqrt{|abcd|})^2 $

$ \frac{1}{16} \ge |abcd| $

This inequality means the absolute value of the product $abcd$ cannot exceed $\frac{1}{16}$.

The maximum value occurs when equality holds in the AM-GM inequality, which happens when $a^2 = b^2 = c^2 = d^2$.

From the constraint $a^2 + b^2 + c^2 + d^2 = 1$, we have $4a^2 = 1$, so $a^2 = \frac{1}{4}$.

This implies $a, b, c, d$ could be $\pm \frac{1}{2}$. To maximize the product $abcd$, we can choose $a = b = c = d = \frac{1}{2}$ (or any combination with an even number of negative signs).

$ abcd = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^4 = \frac{1}{16} $

Thus, the maximum value of the product $abcd$ is $\frac{1}{16}$.