A correlation matrix problem 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: 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
$$\alpha_-=\frac{\sqrt{-\rho_1^4-2 \rho_1^2 \rho_3^2+\rho_2^4-2 \rho_2^3+\rho_2^2-\rho_3^4}+\rho_2^2-\rho_2}{\rho_1^2+\rho_3^2}$$
A: Not a proof, but from a numerical experiment (considering many random correlation matrices) I believe that the statement you propose is true.  The experiment seems to indicate that a sharp bound is given by
$$
\begin{pmatrix}
  \rho_1 & \rho_3
\end{pmatrix}
\begin{pmatrix}
1 & \rho_2
\\
\rho_2  & 1
\end{pmatrix}^{-1}
\begin{pmatrix}
\rho_1 \\ \rho_3
\end{pmatrix}
\le
(1-\alpha)^2,
$$
which implies your bound since $\lambda_\mathrm{min} \in [0,1]$.
Here is the R code I used for my experiment:
n <- 10000
res <- matrix(0, n, 2)
for (i in 1:n) {
  repeat {
    q <- runif(3, -1, 1)
    A <- matrix(c(1,q[1],q[2],q[1],1,q[3],q[2],q[3],1), 3, 3)
    alpha <- min(eigen(A, symmetric = TRUE, only.values = TRUE)$values)
    if (alpha >= 0) {
      break
    }
  }
  B <- matrix(c(1, q[2], q[2], 1), 2, 2)
  x <- c(q[1], q[3])
  beta <- x %*% solve(B, x)
  res[i,] <- c(alpha, beta)
}
plot(res[,1],res[,2], asp=1, cex=.5)
abline(1, -1)
curve((1-x)^2, add = TRUE)

