I have a linear algebraic question about an arbitrary correlation matrix. If


$$
\lambda\_{\min}
\begin{pmatrix}
1 & \rho\_1 & \rho\_2
\\
\rho\_1 & 1 & \rho\_3
\\
\rho\_2 & \rho\_3 & 1
\end{pmatrix}
\ge
\alpha
$$


does the following hold?


$$
(\rho\_1,\rho\_3)
\begin{pmatrix}
1 & \rho\_2
\\
\rho\_2 & 1
\end{pmatrix}^{-1}
\begin{pmatrix}
\rho\_1 \\ \rho\_3
\end{pmatrix}
\le
1-\alpha
$$


Thanks so much.



A correlation matrix problem
If $\rho\_1=\rho\_3=0$ it is obvious, so we assume $\rho\_1^2+\rho\_3^2>0$.
First note that 
$$ A= \begin{pmatrix} 1 & \rho\_2 \\ \rho\_2 & 1 \end{pmatrix}^{-1}=\frac{1}{\rho\_2^2-1}\begin{pmatrix} -1 & \rho\_2 \\ \rho\_2 & -1 \end{pmatrix}$$ with eigenvalues $\lambda\_1=\frac{1}{1-\rho\_2}$ and $\lambda\_2=\frac{1}{1+\rho\_2}$. Since $A$ is symmetric, we know that the Rayleigh quotient of $A$ is upper bounded by $\max\{\lambda\_1,\lambda\_2\}$, that is
$$\frac{\langle Ax,x\rangle}{\langle x,x\rangle} \leq \max\{\lambda\_1,\lambda\_2\} \qquad \forall x\neq 0.$$
In particular for $z=(\rho\_1,\rho\_3)^\top$ we obtain the relation
$$\langle Az,z\rangle \leq \max\{\lambda\_1,\lambda\_2\}(\rho\_1^2+\rho\_3^2)=\max\Big\{\frac{\rho\_1^2+\rho\_3^2}{1-\rho\_2},\frac{\rho\_1^2+\rho\_3^2}{1+\rho\_2}\Big\}.$$
So, we need to prove that 
$$\max\Big\{\frac{\rho\_1^2+\rho\_3^2}{1-\rho\_2},\frac{\rho\_1^2+\rho\_3^2}{1+\rho\_2}\Big\}\leq 1-\alpha 
\iff \min\Big\{1-\frac{\rho\_1^2+\rho\_3^2}{1-\rho\_2},1-\frac{\rho\_1^2+\rho\_3^2}{1+\rho\_2}\Big\}\geq \alpha \tag{1}
$$


Now, the matrix 
$$M =\begin{pmatrix} 1 &\rho\_1 &\rho\_2 \\ \rho\_1&1 & \rho\_3 \\ \rho\_2 & \rho\_3 &1 \end{pmatrix}$$ is also symmetric and thus, again by Rayleigh quotient argument we have
$$\frac{\langle Mu,u\rangle }{\langle u,u\rangle}\geq \alpha \qquad \forall u\neq 0.\tag{2}$$
The idea is now to prove (1) by plugging in (2) some smart choice of $u$. These choices exist but I could not find nicely elegant ones. Here are some choices obtained with Mathematica: Take $u\_{\pm} = (\alpha\_{\pm},0,1)$ with 
$$\alpha\_+=\frac{\sqrt{4 \left(\rho\_2^2+\rho\_2\right)^2-4 \left(\rho\_1^2+\rho\_3^2\right)^2}-2 \rho\_2^2-2 \rho\_2}{2 \left(\rho\_1^2+\rho\_3^2\right)}$$
and
$$\alpha\_-=\frac{\sqrt{-\rho\_1^4-2 \rho\_1^2 \rho\_3^2+\rho\_2^4-2 \rho\_2^3+\rho\_2^2-\rho\_3^4}+\rho\_2^2-\rho\_2}{\rho\_1^2+\rho\_3^2}$$


