To find the maximum value of \( a^4 b^2 c^2 \) given the constraint \( 2a + b + 3c = 1 \) where \( a, b, c \) are positive real numbers, we will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.
### Step-by-step Solution:
1. **Rewrite the constraint**:
We have the equation \( 2a + b + 3c = 1 \). We will use this constraint while applying the AM-GM inequality.
2. **Set up the terms for AM-GM**:
We need to express \( a^4 b^2 c^2 \) in a form suitable for AM-GM.
- We can write \( 2a \) as \( a + a \) (two terms of \( a \)).
- We can write \( b \) as \( b \) (one term of \( b \)).
- We can write \( 3c \) as \( c + c + c \) (three terms of \( c \)).
Thus, we can express the terms as:
\[
a, a, b, c, c, c
\]
3. **Count the number of terms**:
We have a total of 6 terms: \( a, a, b, c, c, c \).
4. **Apply AM-GM inequality**:
According to the AM-GM inequality:
\[
\frac{a + a + b + c + c + c}{6} \geq \sqrt[6]{a^2 b c^3}
\]
The left-hand side simplifies to:
\[
\frac{2a + b + 3c}{6} = \frac{1}{6} \quad \text{(from the constraint)}
\]
Therefore, we have:
\[
\frac{1}{6} \geq \sqrt[6]{a^2 b c^3}
\]
5. **Raise both sides to the power of 6**:
\[
\left(\frac{1}{6}\right)^6 \geq a^2 b c^3
\]
6. **Express \( a^4 b^2 c^2 \)**:
We want to find \( a^4 b^2 c^2 \). Notice that:
\[
a^4 b^2 c^2 = (a^2 b c^3) \cdot \frac{a^2 b}{c}
\]
To maximize \( a^4 b^2 c^2 \), we need to maximize \( a^2 b c^3 \) and \( \frac{a^2 b}{c} \).
7. **Use the constraint**:
From the earlier steps, we know:
\[
a^2 b c^3 \leq \left(\frac{1}{6}\right)^6
\]
We can maximize \( a^4 b^2 c^2 \) by setting \( c = 1 \) and adjusting \( a \) and \( b \) accordingly, but we need to ensure that \( 2a + b + 3c = 1 \).
8. **Find the maximum**:
We can find the maximum value of \( a^4 b^2 c^2 \) by substituting \( a = \frac{1}{8}, b = \frac{1}{4}, c = \frac{1}{8} \) into the expression:
\[
a^4 b^2 c^2 = \left(\frac{1}{8}\right)^4 \left(\frac{1}{4}\right)^2 \left(\frac{1}{8}\right)^2
\]
Calculate:
\[
= \frac{1}{4096} \cdot \frac{1}{16} \cdot \frac{1}{64} = \frac{1}{4096 \cdot 16 \cdot 64}
\]
9. **Final Calculation**:
The maximum value of \( a^4 b^2 c^2 \) is:
\[
= \frac{1}{4096 \cdot 16 \cdot 64} = \frac{1}{4194304}
\]
### Conclusion:
The maximum value of \( a^4 b^2 c^2 \) given the constraint \( 2a + b + 3c = 1 \) is \( \frac{1}{4194304} \).
