One odd pair among four

Idea. A correlation matrix is valid exactly when all its eigenvalues are nonnegative. The symmetries ($a\leftrightarrow d$, $b\leftrightarrow c$) split the $4\times 4$ problem into small pieces.

Step 1: the matrix. In the order $a,b,c,d$: $$M=\left(\begin{array}{cccc}1& 0.5& 0.5& \rho \\ 0.5& 1& 0.5& 0.5\\ 0.5& 0.5& 1& 0.5\\ \rho & 0.5& 0.5& 1\end{array}\right).$$

Step 2: two easy eigenvectors. $M(1,0,0,-1)=(1-\rho )(1,0,0,-1)$, eigenvalue $1-\rho$. $M(0,1,-1,0)=\frac{1}{2}(0,1,-1,0)$, eigenvalue $\frac{1}{2}$.

Step 3: the symmetric part. With $e=(1,0,0,1)$ and $g=(0,1,1,0)$: $$Me=(1+\rho )e+g,Mg=e+\frac{3}{2}g.$$ On this plane $M$ acts like the matrix $$\left(\begin{array}{cc}1+\rho & 1\\ 1& 1.5\end{array}\right),$$ whose determinant is $\frac{3}{2}(1+\rho )-1=\frac{1}{2}+\frac{3}{2}\rho$ and whose trace is positive.

Step 4: the condition. All eigenvalues are nonnegative exactly when $\rho \le 1$ and $\frac{1}{2}+\frac{3}{2}\rho \ge 0$, that is $\rho \ge -\frac{1}{3}$.

The trap. Checking only the $3\times 3$ blocks: the block for $a,b,d$ (with correlations $\frac{1}{2},\frac{1}{2},\rho$) allows $\rho$ down to $-\frac{1}{2}$. The fourth variable adds a constraint the smaller blocks do not see.