Let $0 \le a,$ $b,$ $c \le 1.$  Find the maximum value of
\[\sqrt{abc} + \sqrt{(1 - a)(1 - b)(1 - c)}.\]
Since $0 \le c \le 1,$ $\sqrt{c} \le 1$ and $\sqrt{1 - c} \le 1,$ so
\[\sqrt{abc} + \sqrt{(1 - a)(1 - b)(1 - c)} \le \sqrt{ab} + \sqrt{(1 - a)(1 - b)}.\]Then by AM-GM,
\[\sqrt{ab} \le \frac{a + b}{2}\]and
\[\sqrt{(1 - a)(1 - b)} \le \frac{(1 - a) + (1 - b)}{2} = \frac{2 - a - b}{2},\]so
\[\sqrt{ab} + \sqrt{(1 - a)(1 - b)} \le \frac{a + b}{2} + \frac{2 - a - b}{2} = 1.\]Equality occurs when $a = b = c = 0,$ so the maximum value is $\boxed{1}.$
1
Level 4
Intermediate Algebra
Problem: Let $0 \le a,$ $b,$ $c \le 1.$  Find the maximum value of
\[\sqrt{abc} + \sqrt{(1 - a)(1 - b)(1 - c)}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the maximum value of the expression, we need to analyze the components involved. The expression consists of two parts: the geometric mean of the variables \(a\), \(b\), and \(c\) represented by \(\sqrt{abc}\), and the geometric mean of their complements, \(\sqrt{(1 - a)(1 - b)(1 - c)}\). 

We recognize that both \(abc\) and \((1 - a)(1 - b)(1 - c)\) are non-negative since \(0 \le a, b, c \le 1\). The maximum value of the sum of these two terms will occur when both terms are maximized simultaneously.

To explore the relationship between \(abc\) and \((1 - a)(1 - b)(1 - c)\), we can apply the AM-GM inequality, which states that the arithmetic mean is greater than or equal to the geometric mean. This suggests that the maximum of the sum occurs when the two terms are balanced.

Let us denote \(x = \sqrt{abc}\) and \(y = \sqrt{(1 - a)(1 - b)(1 - c)}\). We want to maximize \(x + y\). The maximum value of \(x\) occurs when \(a = b = c\), and similarly for \(y\) when \(1 - a = 1 - b = 1 - c\). 

Setting \(a = b = c = t\) leads to \(x = \sqrt{t^3}\) and \(y = \sqrt{(1 - t)^3}\). Thus, we need to maximize the function \(f(t) = \sqrt{t^3} + \sqrt{(1 - t)^3}\) for \(0 \le t \le 1\).

To find the critical points, we differentiate \(f(t)\) with respect to \(t\) and set the derivative to zero. The derivative is given by:

\[
f'(t) = \frac{3}{2} t^{1/2} (1 - t)^{3/2} - \frac{3}{2} (1 - t)^{1/2} t^{3/2}
\]

Setting \(f'(t) = 0\) leads to the condition \(t^{1/2} (1 - t)^{3/2} = (1 - t)^{1/2} t^{3/2}\). Simplifying this gives \(t = 1 - t\), leading to \(t = \frac{1}{2}\).

Evaluating \(f\) at \(t = \frac{1}{2}\):

\[
f\left(\frac{1}{2}\right) = \sqrt{\left(\frac{1}{2}\right)^3} + \sqrt{\left(1 - \frac{1}{2}\right)^3} = \sqrt{\frac{1}{8}} + \sqrt{\frac{1}{8}} = 2 \cdot \frac{1}{2\sqrt{2}} = \frac{1}{\sqrt{2}}.
\]

Next, we check the endpoints \(t = 0\) and \(t = 1\):

- At \(t = 0\), \(f(0) = 0 + 1 = 1\).
- At \(t = 1\), \(f(1) = 1 + 0 = 1\).

Thus, the maximum value of \(f(t)\) occurs at both endpoints, yielding a maximum value of \(1\). 

Therefore, the final answer is \(1\).<|END_OF_LATENT|> The final answer is $1$.