Let $a$, $b$, $c$ be the sides of an acute triangle $\triangle ABC$ , then for any $x, y, z \geq 0$, such that $xy+yz+zx=1$ holds inequality:$$a^2x + b^2y + c^2z \geq 4F$$ where $F$ is the area of the triangle $\triangle ABC$
1. **Restate the problem and given conditions:**
   Let \( a, b, c \) be the sides of an acute triangle \(\triangle ABC\). For any \( x, y, z \geq 0 \) such that \( xy + yz + zx = 1 \), we need to prove the inequality:
   \[
   a^2x + b^2y + c^2z \geq 4F
   \]
   where \( F \) is the area of the triangle \(\triangle ABC\).

2. **Express the area \( F \) using the sine theorem:**
   \[
   F = \frac{1}{2}bc \sin A = \frac{1}{2}ab \sin C = \frac{1}{2}ac \sin B
   \]
   Using the sine theorem, we have:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R
   \]
   where \( R \) is the circumradius of the triangle.

3. **Express \( a^2, b^2, c^2 \) in terms of \( F \) and the sines of the angles:**
   \[
   a^2 = 4R^2 \sin^2 A, \quad b^2 = 4R^2 \sin^2 B, \quad c^2 = 4R^2 \sin^2 C
   \]
   Since \( F = \frac{abc}{4R} \), we can write:
   \[
   a^2 = 2F \cdot \frac{\sin A}{\sin B \sin C}, \quad b^2 = 2F \cdot \frac{\sin B}{\sin A \sin C}, \quad c^2 = 2F \cdot \frac{\sin C}{\sin A \sin B}
   \]

4. **Substitute these expressions into the inequality:**
   \[
   a^2x + b^2y + c^2z \geq 4F
   \]
   becomes:
   \[
   2F \left( \frac{x \sin A}{\sin B \sin C} + \frac{y \sin B}{\sin A \sin C} + \frac{z \sin C}{\sin A \sin B} \right) \geq 4F
   \]
   Simplifying, we get:
   \[
   \frac{x \sin A}{\sin B \sin C} + \frac{y \sin B}{\sin A \sin C} + \frac{z \sin C}{\sin A \sin B} \geq 2
   \]

5. **Introduce a new variable \( E \):**
   Let:
   \[
   E = x \sin^2 A + y \sin^2 B + z \sin^2 C
   \]
   We need to show:
   \[
   E \geq 2 \sin A \sin B \sin C
   \]

6. **Case analysis:**
   - **Case 1: \( z = 0 \)**
     \[
     xy = 1 \implies x > 0, y = \frac{1}{x}
     \]
     Using AM-GM inequality:
     \[
     E = x \sin^2 A + \frac{\sin^2 B}{x} \geq 2 \sin A \sin B \geq 2 \sin A \sin B \sin C
     \]
     Equality holds when \( \sin C = 1 \implies C = 90^\circ \) and \( x \sin^2 A = \frac{\sin^2 B}{x} \implies x = \frac{\sin B}{\sin A} = \cot A \).

   - **Case 2: \( z > 0 \)**
     We need to minimize \( E \) under the constraint \( xy + yz + zx = 1 \). Let:
     \[
     x = \frac{1 - yz}{y + z}
     \]
     Then:
     \[
     E(y) = \frac{1 - yz}{y + z} \sin^2 A + y \sin^2 B + z \sin^2 C
     \]
     Taking the derivative and setting it to zero:
     \[
     E'(y) = \frac{(y + z)^2 \sin^2 B - (z^2 + 1) \sin^2 A}{(y + z)^2} = 0
     \]
     Solving for \( y \):
     \[
     y_1 = -\frac{\sin A}{\sin B} \sqrt{z^2 + 1} - z < 0, \quad y_2 = \frac{\sin A}{\sin B} \sqrt{z^2 + 1} - z > y_1
     \]
     Given \( 0 \leq yz \leq 1 \implies 0 \leq y \leq \frac{1}{z} \):
     - If \( y_2 \leq 0 \), the minimum occurs at \( y = 0 \).
     - If \( 0 < y_2 \leq \frac{1}{z} \), the minimum occurs at \( y = y_2 \).
     - If \( y_2 \geq \frac{1}{z} \), the minimum occurs at \( y = \frac{1}{z} \).

     For each sub-case, we verify that:
     \[
     E(y) \geq 2 \sin A \sin B \sin C
     \]

7. **Conclusion:**
   The inequality holds for any triangle, not necessarily acute. The equality occurs for the cumulative conditions:
   \[
   \begin{cases}
   \max\{A, B, C\} \leq 90^\circ \\
   (x, y, z) = (\cot A, \cot B, \cot C)
   \end{cases}
   \]

The final answer is \( \boxed{ a^2x + b^2y + c^2z \geq 4F } \)