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 $$\alpha_-=\frac{\sqrt{-\rho_1^4-2 \rho_1^2 \rho_3^2+\rho_2^4-2 \rho_2^3+\rho_2^2-\rho_3^4}+\rho_2^2-\rho_2}{\rho_1^2+\rho_3^2}$$