Find the smallest volume bounded by the coordinate planes and  by a tangent plane to the ellipsoid
$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1.$$
1. Consider the ellipsoid given by the equation:
   \[
   \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1
   \]
   We need to find the smallest volume bounded by the coordinate planes and a tangent plane to this ellipsoid.

2. Let \((x_0, y_0, z_0)\) be a point on the ellipsoid. The equation of the tangent plane at this point can be derived using the gradient of the ellipsoid function:
   \[
   \nabla \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \right) = \left( \frac{2x}{a^2}, \frac{2y}{b^2}, \frac{2z}{c^2} \right)
   \]
   The equation of the tangent plane at \((x_0, y_0, z_0)\) is:
   \[
   \frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} + \frac{z_0 z}{c^2} = 1
   \]

3. The tangent plane intersects the coordinate axes at:
   \[
   x = \frac{a^2}{x_0}, \quad y = \frac{b^2}{y_0}, \quad z = \frac{c^2}{z_0}
   \]

4. These intercepts form a triangular pyramid (tetrahedron) with the coordinate planes. The volume \(V\) of this tetrahedron is given by:
   \[
   V = \frac{1}{6} \times \text{base area} \times \text{height}
   \]
   The base is a right triangle with sides \(\frac{a^2}{x_0}\) and \(\frac{b^2}{y_0}\), so its area is:
   \[
   \text{Base area} = \frac{1}{2} \times \frac{a^2}{x_0} \times \frac{b^2}{y_0} = \frac{a^2 b^2}{2 x_0 y_0}
   \]
   The height of the tetrahedron is \(\frac{c^2}{z_0}\). Therefore, the volume is:
   \[
   V = \frac{1}{6} \times \frac{a^2 b^2}{2 x_0 y_0} \times \frac{c^2}{z_0} = \frac{a^2 b^2 c^2}{6 x_0 y_0 z_0}
   \]

5. To minimize this volume, we use the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
   \[
   \frac{x_0 y_0 z_0}{abc} \leq \left( \frac{\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} + \frac{z_0^2}{c^2}}{3} \right)^{3/2}
   \]
   Since \(\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} + \frac{z_0^2}{c^2} = 1\), we have:
   \[
   \frac{x_0 y_0 z_0}{abc} \leq \left( \frac{1}{3} \right)^{3/2} = \frac{1}{3\sqrt{3}}
   \]

6. Equality in AM-GM holds when:
   \[
   \frac{x_0}{a} = \frac{y_0}{b} = \frac{z_0}{c} = \frac{1}{\sqrt{3}}
   \]
   Therefore, \(x_0 = \frac{a}{\sqrt{3}}, y_0 = \frac{b}{\sqrt{3}}, z_0 = \frac{c}{\sqrt{3}}\).

7. Substituting these values into the volume formula:
   \[
   V = \frac{a^2 b^2 c^2}{6 x_0 y_0 z_0} = \frac{a^2 b^2 c^2}{6 \left( \frac{a}{\sqrt{3}} \right) \left( \frac{b}{\sqrt{3}} \right) \left( \frac{c}{\sqrt{3}} \right)} = \frac{a^2 b^2 c^2}{6 \frac{abc}{3\sqrt{3}}} = \frac{3\sqrt{3} abc}{6} = \frac{\sqrt{3}}{2} abc
   \]

8. Therefore, the minimum volume is:
   \[
   \boxed{\frac{\sqrt{3}}{2} abc}
   \]