To prove that \( \frac{1}{x^2+y^2} + \frac{3}{2xy} \geq 2 \) for \( x, y > 0 \) and \( x + y = 2 \), we start with the given conditions and manipulate the expressions.

Since \( x + y = 2 \), we can express \( y \) in terms of \( x \):
\[
y = 2 - x.
\]

Now we can express \( x^2 + y^2 \) and \( xy \):
\[
x^2 + y^2 = x^2 + (2 - x)^2 = x^2 + (4 - 4x + x^2) = 2x^2 - 4x + 4.
\]
Combining like terms gives us:
\[
x^2 + y^2 = 2(x^2 - 2x + 2) = 2(x - 1)^2 + 2.
\]

Next, we compute \( xy \):
\[
xy = x(2 - x) = 2x - x^2.
\]

We know the relationship between \( x^2 + y^2 \) and \( xy \):
\[
x^2 + y^2 = (x+y)^2 - 2xy = 2^2 - 2xy = 4 - 2xy.
\]

Therefore, we express the sum:
\[
\frac{1}{x^2 + y^2} + \frac{3}{2xy} = \frac{1}{4 - 2xy} + \frac{3}{2xy}.
\]

Let \( z = xy \), leading to:
\[
\frac{1}{4 - 2z} + \frac{3}{2z}.
\]

We need to find the minimum of this function knowing that \( z \) can vary depending on \( x \) and \( y \) under the constraint \( x + y = 2 \). The maximum value of \( z \) occurs when \( x = y = 1 \):
\[
xy \leq 1 \implies z \leq 1.
\]

Thus, the range of \( z \) is \( 0 < z \leq 1 \).

Now we need to show:
\[
\frac{1}{4 - 2z} + \frac{3}{2z} \geq 2.
\]
To do this, let's bring the terms to a common denominator:
\[
\frac{2z + 6 - 4z}{(4 - 2z)(2z)} = \frac{6 - 2z}{(4 - 2z)(2z)}.
\]
We need to show that:
\[
6 - 2z \geq 2(4 - 2z)(2z).
\]

This can be simplified further. We multiply both sides by \( (4 - 2z)(2z) \) (which is always positive since \( z > 0 \)):
\[
6 - 2z \geq 8z - 8z^2.
\]
Rearranging terms gives us:
\[
8z^2 - 10z + 6 \geq 0.
\]

Solving this quadratic inequality, we want to find the roots:
\[
z = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 8 \cdot 6}}{2 \cdot 8} = \frac{10 \pm \sqrt{100 - 192}}{16} = \frac{10 \pm \sqrt{-92}}{16}.
\]
Since the discriminant is negative, the quadratic \( 8z^2 -10z + 6 \) does not cross the x-axis, and we analyze the sign of the quadratic:

The leading coefficient (8) is positive, which means \( 8z^2 - 10z + 6 \) is always positive.

This confirms that:
\[
\frac{1}{x^2+y^2} + \frac{3}{2xy} \geq 2.
\]

Thus, we have shown:
\[
\frac{1}{x^2+y^2} + \frac{3}{2xy} \geq 2.
\]
QED.