### Instruction: Let $a,$ $b,$ and $c$ be nonnegative real numbers such that $a + b + c = 1.$  Find the maximum value of
\[a + \sqrt{ab} + \sqrt[3]{abc}.\].### Response: I notice that this expression involves the arithmetic, geometric, and harmonic means of $a, b,$ and $c.$
I recall that the arithmetic mean is always greater than or equal to the geometric mean, and the geometric mean is always greater than or equal to the harmonic mean.
This means that if I replace each term in the expression with its arithmetic mean, I will get a larger value.
For example, $\sqrt{ab} \leq \frac{a + b}{2}$ and $\sqrt[3]{abc} \leq \frac{a + b + c}{3}.$
So, the expression is less than or equal to $a + \frac{a + b}{2} + \frac{a + b + c}{3}.$
Simplifying, I get $a + \frac{a + b}{2} + \frac{a + b + c}{3} = \frac{11}{6} a + \frac{5}{6} b + \frac{1}{6} c.$
Now, since $a, b,$ and $c$ are nonnegative and add up to $1,$ they must all be between $0$ and $1.$
This means that the coefficients of $a, b,$ and $c$ in this expression are also between $0$ and $1.$
Therefore, the maximum value of this expression occurs when $a = 1$ and $b = c = 0.$
Plugging these values in, I get $\frac{11}{6} a + \frac{5}{6} b + \frac{1}{6} c = \frac{11}{6}.$
So, the maximum value of the original expression is $\frac{11}{6}.$