By Cauchy-Schwarz,
\[(a + b + c + d) \left( \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \right) \ge (1 + 1 + 2 + 4)^2 = 64.\]Equality occurs when $a = b = \frac{c}{2} = \frac{d}{4}$ and $a + b + c + d = 1.$  We can solve to get $a = \frac{1}{8},$ $b = \frac{1}{8},$ $c = \frac{1}{4},$ and $d = \frac{1}{2},$ so the minimum value is $\boxed{64}.$
Question:
Let $a,$ $b,$ $c,$ $d$ be positive real numbers such that $a + b + c + d = 1.$  Find the minimum value of
\[\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}.\]
Answer:
Let's think step by step.

user
64
Let $a,$ $b,$ $c,$ $d$ be positive real numbers such that $a + b + c + d = 1.$  Find the minimum value of
\[\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}.\]
test