In the paper 'On the Holder continuity of solutions of second order elliptic equations in two variables' by Piccinini and Spagnolo, they prove the following estimate:
$$
\begin{array}{ll}
\left(\int\_S p\_{11} |u\_T|^2 \right)^{\frac12}\left(\int\_S \frac{\langle P(x) (u\_N,u\_T), (1,0) \rangle^2}{p\_{11}} \right)^{\frac12} &\leq & \sqrt{\frac{\Lambda}{\lambda}} \left(\int\_S \lambda |u\_T|^2 \right)^{\frac12}\left(\int\_S \frac{\langle P(x) (u\_N,u\_T), (1,0) \rangle^2}{p\_{11}} \right)^{\frac12} \\
& \leq & \sqrt{\frac{\Lambda}{\lambda}}\frac12 \left(\int\_S \lambda |u\_T|^2 + \frac{(p\_{11} u\_N + p\_{12}u\_T)^2}{p\_{11}} \right)\\
& \leq & \sqrt{\frac{\Lambda}{\lambda}}\frac12 \left(\int\_S \left( p\_{22} - \frac{p\_{12}^2}{p\_{11}}\right) |u\_T|^2 + \frac{(p\_{11} u\_N + p\_{12}u\_T)^2}{p\_{11}} \right)\\
& = & \sqrt{\frac{\Lambda}{\lambda}}\frac12 \left(\int\_S \langle P(x) (u\_N,u\_T),(u\_N,u\_T)\rangle \right)
\end{array}
$$
Here $P(x)$ is a symmetric positive definite $2\times2$ matrix with entries $(p\_{ij})$ and $\lambda |\xi|^2 \leq \langle P(x)\xi,\xi\rangle \leq \Lambda |\xi|^2$ and $(u\_N,u\_T)$ are the normal and tangential derivative with $S$ being the unit circle in $\mathbb{R}^2$.


The crucial part of the above calculation is the inequality $\lambda \leq \left( p\_{22} - \frac{p\_{12}^2}{p\_{11}}\right)$.


My question is the following: Is there a different way to prove this estimate without explicitly writing down the expressions and instead proceed using only matrix inequalities?



Matrix inequality in a paper by Piccinini-Spagnolo
Let $0<\lambda\_1\le\lambda\_2$ be the eigenvalues of the symmetric positive $2\times2$ matrix $P$. Then $$\lambda\le\lambda\_1=\min\_{|\xi|=1} \langle P\xi,\xi\rangle \le\langle Pe\_1,e\_1\rangle=p\_{11}\le \lambda\_2 =\max\_{|\xi|=1} \langle P\xi,\xi\rangle,$$
so $\lambda p\_{11}\le \lambda\_1\lambda\_2=\det P=p\_{11}p\_{22}-p\_{12}^2$, whence the inequality, since $\langle Pe\_1,e\_1\rangle=p\_{11}>0$.


