$a,b,c,x,y,z$ are positive real numbers and $bz+cy=a$, $az+cx=b$, $ay+bx=c$. Find the least value of following function
$f(x,y,z)=\frac{x^2}{1+x}+\frac{y^2}{1+y}+\frac{z^2}{1+z}$
1. **Expressing \(x\), \(y\), and \(z\) in terms of \(a\), \(b\), and \(c\):**

   Given the equations:
   \[
   bz + cy = a, \quad az + cx = b, \quad ay + bx = c
   \]
   we can solve for \(x\), \(y\), and \(z\) in terms of \(a\), \(b\), and \(c\). Using the Law of Cosines, we get:
   \[
   x = \frac{-a^2 + b^2 + c^2}{2bc}, \quad y = \frac{a^2 - b^2 + c^2}{2ac}, \quad z = \frac{a^2 + b^2 - c^2}{2ab}
   \]
   If \(x\), \(y\), and \(z\) are positive, then \(a\), \(b\), and \(c\) must be the sides of an acute triangle. By the Law of Cosines, we have:
   \[
   x = \cos A, \quad y = \cos B, \quad z = \cos C
   \]
   where \(A\), \(B\), and \(C\) are the angles opposite sides \(a\), \(b\), and \(c\) respectively.

2. **Sum of cosines:**

   Let \(s = x + y + z\). We claim that \(s \geq \frac{3}{2}\).

   **Proof:**
   Since \(0 < A, B, C < \frac{\pi}{2}\), by Jensen's Inequality applied to the concave function \(\cos\), we have:
   \[
   \cos A + \cos B + \cos C \geq \frac{3}{2}
   \]
   This is because the function \(\cos\) is concave on \((0, \frac{\pi}{2})\).

3. **Applying Cauchy-Schwarz Inequality:**

   Using the Cauchy-Schwarz Inequality, we get:
   \[
   \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{(x+y+z)^2}{3 + x + y + z}
   \]
   Substituting \(s = x + y + z\), we have:
   \[
   \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{s^2}{3+s}
   \]

4. **Minimizing the function:**

   We need to show that:
   \[
   \frac{s^2}{3+s} \geq \frac{1}{2}
   \]
   This inequality holds if:
   \[
   \frac{s^2}{3+s} \geq \frac{1}{2} \iff 2s^2 \geq 3 + s \iff 2s^2 - s - 3 \geq 0
   \]
   Solving the quadratic inequality:
   \[
   2s^2 - s - 3 = 0 \implies s = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm 5}{4}
   \]
   Thus, \(s = \frac{3}{2}\) or \(s = -1\). Since \(s\) is positive, we have \(s \geq \frac{3}{2}\).

   Therefore:
   \[
   \frac{s^2}{3+s} \geq \frac{\left(\frac{3}{2}\right)^2}{3 + \frac{3}{2}} = \frac{\frac{9}{4}}{\frac{9}{2}} = \frac{1}{2}
   \]

   Hence, the minimum value of the function \(f(x, y, z)\) is \(\frac{1}{2}\), achieved when \(x = y = z = \frac{1}{2}\).

The final answer is \(\boxed{\frac{1}{2}}\)