Find positive reals $a, b, c$ which maximizes the value of $a+ 2b+ 3c$ subject to the constraint that $9a^2 + 4b^2 + c^2 = 91$
To find the positive reals \(a, b, c\) that maximize the value of \(a + 2b + 3c\) subject to the constraint \(9a^2 + 4b^2 + c^2 = 91\), we can use the method of Lagrange multipliers. However, the given solution uses the Cauchy-Schwarz inequality, which is also a valid approach.

1. **Apply the Cauchy-Schwarz Inequality:**
   The Cauchy-Schwarz inequality states that for any real numbers \(x_1, x_2, \ldots, x_n\) and \(y_1, y_2, \ldots, y_n\),
   \[
   (x_1^2 + x_2^2 + \cdots + x_n^2)(y_1^2 + y_2^2 + \cdots + y_n^2) \geq (x_1y_1 + x_2y_2 + \cdots + x_ny_n)^2.
   \]
   In our case, let \(x_1 = 3a\), \(x_2 = 2b\), \(x_3 = c\), and \(y_1 = \frac{1}{3}\), \(y_2 = 1\), \(y_3 = 3\). Then,
   \[
   (9a^2 + 4b^2 + c^2)\left(\left(\frac{1}{3}\right)^2 + 1^2 + 3^2\right) \geq (a \cdot \frac{1}{3} + 2b \cdot 1 + 3c \cdot 3)^2.
   \]

2. **Simplify the Inequality:**
   \[
   (9a^2 + 4b^2 + c^2)\left(\frac{1}{9} + 1 + 9\right) \geq (a + 2b + 3c)^2.
   \]
   \[
   (9a^2 + 4b^2 + c^2) \cdot \frac{91}{9} \geq (a + 2b + 3c)^2.
   \]
   Given the constraint \(9a^2 + 4b^2 + c^2 = 91\), we substitute:
   \[
   91 \cdot \frac{91}{9} \geq (a + 2b + 3c)^2.
   \]
   \[
   \frac{91^2}{9} \geq (a + 2b + 3c)^2.
   \]
   \[
   \frac{8281}{9} \geq (a + 2b + 3c)^2.
   \]
   \[
   \frac{91}{3} \geq a + 2b + 3c.
   \]

3. **Equality Condition:**
   Equality in the Cauchy-Schwarz inequality holds when the vectors are proportional. Thus, we need:
   \[
   \frac{3a}{\frac{1}{3}} = \frac{2b}{1} = \frac{c}{3}.
   \]
   Simplifying these ratios, we get:
   \[
   9a = 2b = \frac{c}{3}.
   \]

4. **Solve for \(a, b, c\):**
   Let \(9a = k\), then \(a = \frac{k}{9}\).
   \[
   2b = k \implies b = \frac{k}{2}.
   \]
   \[
   \frac{c}{3} = k \implies c = 3k.
   \]
   Substitute \(a, b, c\) back into the constraint:
   \[
   9\left(\frac{k}{9}\right)^2 + 4\left(\frac{k}{2}\right)^2 + (3k)^2 = 91.
   \]
   \[
   \frac{k^2}{9} + k^2 + 9k^2 = 91.
   \]
   \[
   \frac{k^2}{9} + 10k^2 = 91.
   \]
   \[
   \frac{91k^2}{9} = 91.
   \]
   \[
   k^2 = 9.
   \]
   \[
   k = 3.
   \]
   Therefore,
   \[
   a = \frac{3}{9} = \frac{1}{3}, \quad b = \frac{3}{2}, \quad c = 9.
   \]

The final answer is \(\boxed{\frac{1}{3} + 2 \cdot \frac{3}{2} + 3 \cdot 9 = \frac{1}{3} + 3 + 27 = 30.333}\).