Balkan Mathematical Olympiad

Solution:
Let us first write the expression $L$ on the left hand side in the following way
$$\begin{aligned} L & = \left(\frac{2x^{2}-x+y+z}{x+y^{2}+z^{2}} + 2\right) + \left(\frac{2y^{2}+x-y+z}{x^{2}+y+z^{2}} + 2\right) + \left(\frac{2z^{2}+x+y-z}{x^{2}+y^{2}+z} + 2\right) - 6 \\ & = \left(2x^{2} + 2y^{2} + 2z^{2} + x + y + z\right)\left(\frac{1}{x+y^{2}+z^{2}} + \frac{1}{x^{2}+y+z^{2}} + \frac{1}{x^{2}+y^{2}+z}\right) - 6 \end{aligned}$$
If we introduce the notation $A = x + y^{2} + z^{2}$, $B = x^{2} + y + z^{2}$, $C = x^{2} + y^{2} + z$, then the previous relation becomes
$$L = (A + B + C)\left(\frac{1}{A} + \frac{1}{B} + \frac{1}{C}\right) - 6$$
Using the arithmetic-harmonic mean inequality or Cauchy-Schwarz inequality for positive real numbers $A, B, C$, we easily obtain
$$(A + B + C)\left(\frac{1}{A} + \frac{1}{B} + \frac{1}{C}\right) \geqslant 9$$
so it holds $L \geqslant 3$.

The equality occurs if and only if $A = B = C$, which is equivalent to the system of equations
$$x^{2} - y^{2} = x - y, \quad y^{2} - z^{2} = y - z, \quad x^{2} - z^{2} = x - z$$
It follows easily that the only solutions of this system are
$(x, y, z) \in \{(t, t, t) \mid t > 0\} \cup \{(t, t, 1-t) \mid t \in [0,1]\} \cup \{(t, 1-t, t) \mid t \in [0,1]\} \cup \{(1-t, t, t) \mid t \in [0,1]\}$.