Okay, so I need to find the minimum value of the expression \(\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}\) given that \(a + b + c + d = 1\) and all variables are positive real numbers. Hmm, minima problems with constraints usually make me think of using the method of Lagrange multipliers, but maybe there's a simpler approach using inequalities like Cauchy-Schwarz or AM-HM. Let me explore both options.

First, let's recall the Cauchy-Schwarz inequality. In its basic form, for sequences of positive numbers, it says that \((\sum u_i v_i)^2 \leq (\sum u_i^2)(\sum v_i^2)\). But here, I need to relate the sum of reciprocals to the sum of the variables. Alternatively, the AM-HM inequality states that \(\frac{x_1 + x_2 + \dots + x_n}{n} \geq \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}}\). However, in this case, our expression is a weighted sum of reciprocals, so maybe a weighted AM-HM approach would be better.

Wait, maybe using the method of Lagrange multipliers is more straightforward here. Let me set up the function to minimize: \(f(a, b, c, d) = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}\), with the constraint \(g(a, b, c, d) = a + b + c + d - 1 = 0\).

The Lagrangian would be \(L = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} - \lambda(a + b + c + d - 1)\).

Taking partial derivatives with respect to each variable and setting them to zero:

1. \(\frac{\partial L}{\partial a} = -\frac{1}{a^2} - \lambda = 0\)
2. \(\frac{\partial L}{\partial b} = -\frac{1}{b^2} - \lambda = 0\)
3. \(\frac{\partial L}{\partial c} = -\frac{4}{c^2} - \lambda = 0\)
4. \(\frac{\partial L}{\partial d} = -\frac{16}{d^2} - \lambda = 0\)
5. The constraint: \(a + b + c + d = 1\)

From the first four equations, we can express \(\lambda\) in terms of each variable:

From equation 1: \(\lambda = -\frac{1}{a^2}\)

From equation 2: \(\lambda = -\frac{1}{b^2}\)

From equation 3: \(\lambda = -\frac{4}{c^2}\)

From equation 4: \(\lambda = -\frac{16}{d^2}\)

Since all these expressions equal \(\lambda\), we can set them equal to each other:

\(-\frac{1}{a^2} = -\frac{1}{b^2}\) implies \(a = b\)

Similarly, \(-\frac{1}{a^2} = -\frac{4}{c^2}\) implies \(\frac{1}{a^2} = \frac{4}{c^2}\) so \(c = 2a\)

And \(-\frac{1}{a^2} = -\frac{16}{d^2}\) implies \(\frac{1}{a^2} = \frac{16}{d^2}\) so \(d = 4a\)

Therefore, we can express all variables in terms of \(a\):

\(b = a\), \(c = 2a\), \(d = 4a\)

Now, substitute these into the constraint equation:

\(a + a + 2a + 4a = 1\)

Adding them up: \(8a = 1\) => \(a = \frac{1}{8}\)

Then, substituting back:

\(b = \frac{1}{8}\), \(c = 2 \times \frac{1}{8} = \frac{1}{4}\), \(d = 4 \times \frac{1}{8} = \frac{1}{2}\)

Now, compute the original expression:

\(\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}\)

Plugging in the values:

\(\frac{1}{\frac{1}{8}} + \frac{1}{\frac{1}{8}} + \frac{4}{\frac{1}{4}} + \frac{16}{\frac{1}{2}}\)

Calculating each term:

\(8 + 8 + 16 + 32 = 64\)

So the minimum value is 64. Wait, that seems straightforward, but let me check if I did everything correctly.

First, checking the derivatives. The partial derivatives with respect to each variable minus lambda equal zero. So we have:

For a: \( -1/a^2 - \lambda = 0 \implies \lambda = -1/a^2 \)

For b: same as a, so \( \lambda = -1/b^2 \implies a = b \)

For c: \( -4/c^2 - \lambda = 0 \implies \lambda = -4/c^2 \implies -1/a^2 = -4/c^2 \implies c^2 = 4a^2 \implies c = 2a \)

Similarly, for d: \( -16/d^2 - \lambda = 0 \implies \lambda = -16/d^2 \implies -1/a^2 = -16/d^2 \implies d^2 = 16a^2 \implies d = 4a \)

So all substitutions are correct. Then adding up a + a + 2a + 4a = 8a = 1 => a = 1/8. Then substituting back gives c = 1/4, d = 1/2, which are correct. Then plugging into the expression:

1/(1/8) = 8

1/(1/8) = 8

4/(1/4) = 16

16/(1/2) = 32

Adding them up: 8 + 8 + 16 + 32 = 64. That's correct.

Alternatively, maybe using Cauchy-Schwarz or AM-HM. Let me see if there's another way to approach this.

Suppose we consider the expression as a sum of terms each involving 1/a, 1/b, 4/c, 16/d. Maybe we can use Cauchy-Schwarz on these terms.

Let me think. Let's think of the expression as:

(1/a + 1/b + 4/c + 16/d) * (a + b + c + d) >= (1 + 1 + 2 + 4)^2

Wait, let's apply Cauchy-Schwarz in the form:

(\sum \frac{1}{x_i}) (\sum x_i) >= (\sum 1)^2

But here, the weights are different. The standard Cauchy-Schwarz would be:

(\sum (1/\sqrt{x_i})^2)(\sum (\sqrt{x_i})^2) >= (\sum 1)^2

But in our case, the terms are not all the same. Hmm.

Alternatively, maybe use the method of weighted AM-HM inequality. The weighted AM-HM states that:

\(\frac{\sum w_i x_i}{\sum w_i} \geq \frac{\sum w_i}{\sum \frac{w_i}{x_i}}\)

But not sure. Alternatively, maybe use the method of Lagrange multipliers as above is the most straightforward here.

Alternatively, let's consider variables scaled by their coefficients. Let me set up variables such that each term in the expression has the same denominator. For example, let me write:

Let’s denote \(x = a\), \(y = b\), \(z = c\), \(w = d\). Then the expression is \(1/x + 1/y + 4/z + 16/w\) and the constraint is \(x + y + z + w = 1\).

Alternatively, maybe use substitution. Let me set \(u = a\), \(v = b\), \(m = c\), \(n = d\). Then the same as above.

Alternatively, suppose we use variables substitution where we set each term's coefficient to be squared or something. Wait, maybe not. Alternatively, think of the coefficients as squares: 1, 1, 2^2, 4^2. Wait, 1,1,4,16. Wait, 1,1,4,16 are 1^2, 1^2, 2^2, 4^2. Hmm, but not sure.

Alternatively, maybe use the Cauchy-Schwarz inequality in the following way:

Let’s consider vectors:

Let’s have vectors \( (1/\sqrt{a}, 1/\sqrt{b}, 2/\sqrt{c}, 4/\sqrt{d}) \) and \( (\sqrt{a}, \sqrt{b}, \sqrt{c}/2, \sqrt{d}/4) \). Then their dot product is:

\( (1/\sqrt{a} \cdot \sqrt{a}) + (1/\sqrt{b} \cdot \sqrt{b}) + (2/\sqrt{c} \cdot \sqrt{c}/2) + (4/\sqrt{d} \cdot \sqrt{d}/4) = 1 + 1 + 1 + 1 = 4 \)

By Cauchy-Schwarz, the square of this dot product is less than or equal to the product of the sums of squares of each vector.

So:

\( (4)^2 \leq \left( \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \right) \left( a + b + \frac{c}{4} + \frac{d}{16} \right) \)

But our constraint is \(a + b + c + d = 1\). However, the second term in the product is \(a + b + c/4 + d/16\). To relate this to the given constraint, we need to make sure that \(c/4 + d/16 \leq c + d\). Which is true since \(c/4 \leq c\) and \(d/16 \leq d\) for positive c, d. But since we have \(a + b + c + d = 1\), maybe we can bound the second term.

But perhaps not helpful here. Alternatively, maybe we can adjust variables. Let’s set \(c' = c/4\) and \(d' = d/16\), then \(c = 4c'\), \(d = 16d'\). Then the constraint becomes \(a + b + 4c' + 16d' = 1\). But the expression becomes \(\frac{1}{a} + \frac{1}{b} + \frac{4}{4c'} + \frac{16}{16d'} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c'} + \frac{1}{d'}\). So now, we have to minimize \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c'} + \frac{1}{d'}\) with \(a + b + 4c' + 16d' = 1\). But this might complicate further.

Alternatively, maybe use AM-HM on the terms. Let's consider the terms as four separate terms:

But in the original expression, we have four terms: 1/a, 1/b, 4/c, 16/d. Let's consider using Cauchy-Schwarz with weights. The Cauchy-Schwarz inequality in weighted form states that:

\(\left( \sum_{i=1}^n w_i x_i \right) \left( \sum_{i=1}^n \frac{x_i}{w_i} \right) \geq \left( \sum_{i=1}^n x_i \right)^2\)

But in our case, if we set \(w_1 = 1\), \(w_2 = 1\), \(w_3 = 2\), \(w_4 = 4\), and \(x_1 = 1/a\), \(x_2 = 1/b\), \(x_3 = 4/c\), \(x_4 = 16/d\), then:

\(\left( 1 \cdot \frac{1}{a} + 1 \cdot \frac{1}{b} + 2 \cdot \frac{4}{c} + 4 \cdot \frac{16}{d} \right) \left( 1 \cdot a + 1 \cdot b + \frac{1}{2} \cdot c + \frac{1}{4} \cdot d \right) \geq (1 + 1 + 2 + 4)^2 = 36\)

But the left side is \(\left( \frac{1}{a} + \frac{1}{b} + \frac{8}{c} + \frac{64}{d} \right) \left( a + b + \frac{c}{2} + \frac{d}{4} \right)\), which is different from our original expression. So maybe not useful here.

Alternatively, maybe use Hölder's inequality? Hölder's inequality generalizes Cauchy-Schwarz and might be applicable here. Hölder's inequality states that for positive real numbers and exponents p, q such that 1/p + 1/q = 1:

\(\sum a_i b_i \leq (\sum a_i^p)^{1/p} (\sum b_i^q)^{1/q}\)

But again, not sure how to apply it here.

Alternatively, perhaps use the AM-GM inequality on the terms. Since we have the sum of reciprocals, AM-GM might give a lower bound. Let's see.

The AM-GM inequality states that for positive real numbers, the arithmetic mean is at least the geometric mean. So, for four terms:

\(\frac{\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}}{4} \geq \sqrt[4]{\frac{1}{a} \cdot \frac{1}{b} \cdot \frac{4}{c} \cdot \frac{16}{d}}\)

Calculating the geometric mean:

\(\sqrt[4]{\frac{1 \cdot 1 \cdot 4 \cdot 16}{a b c d}} = \sqrt[4]{\frac{64}{a b c d}}\)

But since \(a + b + c + d = 1\), the product \(a b c d\) is maximized when all variables are equal, but here we have different coefficients. However, the AM-GM approach here gives a lower bound of \(\sqrt[4]{64/(a b c d)}\), but this doesn't directly give a numerical value. To use this, we would need to find a relationship between the product \(a b c d\) and the sum \(a + b + c + d\), which might not be straightforward.

Alternatively, perhaps use the method of substitution. Let me set \(x = a\), \(y = b\), \(z = c\), \(w = d\) again. Then we need to minimize \(1/x + 1/y + 4/z + 16/w\) under \(x + y + z + w = 1\).

Suppose we fix the ratios between variables. For example, from the Lagrange multiplier method, we found that \(a = b\), \(c = 2a\), \(d = 4a\). Then substituting into the constraint gives \(a + a + 2a + 4a = 8a = 1\), so \(a = 1/8\). Then the variables are determined as above, leading to the minimal value of 64. This seems straightforward, and since the method of Lagrange multipliers gives a consistent solution, it's likely correct.

Alternatively, let's test if this is indeed the minimum by checking another point. Suppose we take different values for \(a, b, c, d\) that sum to 1 and compute the expression.

For example, let’s take \(a = 1/4\), \(b = 1/4\), \(c = 1/4\), \(d = 1/4\). Then the expression becomes:

\(4 + 4 + 16 + 64 = 88\), which is higher than 64.

Another test case: \(a = 1/2\), \(b = 1/2\), \(c = 0\), \(d = 0\). But c and d must be positive, so this is invalid. Let's take \(a = 1/3\), \(b = 1/3\), \(c = 1/3\), \(d = 0\), but again d must be positive. Let's take \(a = 1/3\), \(b = 1/3\), \(c = 1/6\), \(d = 1/6\). Then the expression is:

\(3 + 3 + 4/(1/6) + 16/(1/6) = 3 + 3 + 24 + 96 = 126\), which is much higher.

Alternatively, take \(a = 1/5\), \(b = 1/5\), \(c = 2/5\), \(d = 0\). But d must be positive. Let's take \(a = 1/5\), \(b = 1/5\), \(c = 1/5\), \(d = 3/5\). Then the expression is:

\(5 + 5 + 4/(1/5) + 16/(3/5) = 5 + 5 + 20 + 80/3 ≈ 5 + 5 + 20 + 26.666 ≈ 56.666\), which is still higher than 64.

Alternatively, take \(a = 1/8\), \(b = 1/8\), \(c = 1/4\), \(d = 1/2\), which is the solution we found. The expression is 64. So, another test case near this point: \(a = 1/8 + \epsilon\), \(b = 1/8 - \epsilon\), \(c = 1/4\), \(d = 1/2\). Let's compute the expression:

\(\frac{1}{1/8 + \epsilon} + \frac{1}{1/8 - \epsilon} + \frac{4}{1/4} + \frac{16}{1/2}\)

Simplify:

\(\frac{1}{1/8 + \epsilon} + \frac{1}{1/8 - \epsilon} + 16 + 32\)

Which is \(\frac{1}{1/8 + \epsilon} + \frac{1}{1/8 - \epsilon} + 48\)

Compute the first two terms:

Using \(x = 1/8\), the expression becomes \(\frac{1}{x + \epsilon} + \frac{1}{x - \epsilon} = \frac{2x}{x^2 - \epsilon^2}\). For small \(\epsilon\), this is approximately \(\frac{2x}{x^2} = \frac{2}{x} = 16\). Therefore, the total expression is approximately \(16 + 48 = 64\). So, perturbing \(a\) and \(b\) slightly around 1/8 doesn't change the total much, confirming that 64 is indeed the minimum.

Therefore, after considering multiple approaches and verifying with test cases, the minimum value is 64.
Therefore, the minimum value of the given expression is $\boxed{64}$.
</think>