$$\begin{aligned} |a|^2 &\le (|b|^2 + |c|^2)(|z|^2 + |y|^2) \\ |b|^2 &\le (|c|^2 + |a|^2)(|x|^2 + |z|^2) \\ |c|^2 &\le (|b|^2 + |a|^2)(|x|^2 + |y|^2), \end{aligned}$$
whence
$$\frac{|a|^2}{|b|^2 + |c|^2} + \frac{|b|^2}{|c|^2 + |a|^2} + \frac{|c|^2}{|a|^2 + |b|^2} \le 2 (|x|^2 + |y|^2 + |z|^2).$$
By Nesbitt's inequality, the LHS exceeds $3/2$. Hence the result.

If $a = b = c = 1$, then $x = y = z = 1/2$, in which case $x^2 + y^2 + z^2 = 3/4$ and so the inequality is the best possible. Since in any triangle,
$$\begin{aligned} a &= b \cos C + c \cos B, \\ b &= c \cos A + a \cos C, \\ c &= a \cos B + b \cos A, \end{aligned}$$
we obtain the consequence.