Let $0<\lambda_{1} \leq \ldots \leq \lambda_{n}$ be the eigenvalues of the symmetric and positive definite matrix $A \in \mathbb{R}^{n \times > n}$ and $u_{1}, \ldots, u_{n} \in \mathbb{R}^{n}$ corresponding parwise orthonormal eigenvectors. For $x \neq 0$ let $$ F(x):=\frac{\left(x^{\top} x\right)^{2}}{\left(x^{\top} A > x\right)\left(x^{\top} A^{-1} x\right)} \text {. } $$
- show: $$ \begin{array}{l} F(x)=\frac{1}{\bar{\lambda}\left(\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}^{-1}\right)} \\ \text { with } x=\sum \limits_{i=1}^{n} \beta_{i} u_{i}, \beta_{i} \in \mathbb{R}, \gamma_{i}:=\left(\sum \limits_{j=1} \beta_{j}^{2}\right)^{-1} \beta_{i}^{2} \text { and } \bar{\lambda}:=\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i} \text { for } i=1, \ldots, n \text {. }\end{array} $$
- show $F(x) \geq 4 \lambda_{1} \lambda_{n}\left(\lambda_{1}+\lambda_{n}\right)^{-2}$. Where is this statement needed in connection with descent methods?
For the first part I would have thought: $$\begin{array}{l}x^{\top} A x=\left(\sum \limits_{i=1}^{n} \beta_{i} u_{i}\right)^{\top} Q \Lambda Q^{\top}\left(\sum \limits_{i=1}^{n} \beta_{i} u_{i}\right)=\left(\sum \limits_{i=1}^{n} \beta_{i} u_{i}\right)^{\top} \Lambda\left(\sum \limits_{i=1}^{n} \beta_{i} u_{i}\right)=\sum \limits_{i=1}^{n} \lambda_{i} \beta_{i}^{2} \\ x^{\top} A^{-1} x=\sum \limits_{i=1}^{n} \lambda_{i}^{-1} \beta_{i}^{2} \\ \bar{\lambda}=\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}=\sum \limits_{i=1}^{n}\left(\sum \limits_{j=1}^{n} \beta_{j}^{2}\right)^{-1} \beta_{i}^{2} \lambda_{i} \\ \sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}^{-1}=\sum \limits_{i=1}^{n}\left(\sum \limits_{j=1}^{n} \beta_{j}^{2}\right)^{-1} \beta_{i}^{2} \lambda_{i}^{-1} \\ F(x)=\frac{\left(\sum \limits_{i=1}^{n} \beta_{i}^{2}\right)^{2}}{\left(\sum \limits_{i=1}^{n} \lambda_{i} \beta_{i}^{2}\right)\left(\sum \limits_{i=1}^{n} \lambda_{i}^{-1} \beta_{i}^{2}\right)}=\frac{1}{\frac{\left(\sum \limits_{i=1}^{n} \beta_{i}^{2}\right)^{2}}{\left(\sum \limits_{i=1}^{n} \lambda_{i} \beta_{i}^{2}\right)\left(\sum \limits_{i=1}^{n} \lambda_{i}^{-1} \beta_{i}^{2}\right)}}=\frac{1}{\frac{\left(\sum \limits_{i=1}^{n} \beta_{i}^{2}\right)^{2}}{\left(\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}\right)\left(\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}^{-1}\right)}} \\ \left(\sum \limits_{i=1}^{n} \beta_{i}^{2}\right)^{2}=\left(\sum \limits_{i=1}^{n} \beta_{i}^{2}\right) \cdot\left(\sum \limits_{i=1}^{n} \beta_{i}^{2}\right)=\left(\sum \limits_{i=1}^{n} \beta_{i}^{2}\right) \cdot\left(\sum \limits_{i=1}^{n} \gamma_{i} \beta_{i}^{2}\right) \\ F(x)=\frac{1}{\frac{\left(\sum \limits_{i=1}^{n} \beta_{i}^{2}\right)^{2}}{\left(\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}\right)\left(\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}^{-1}\right)}}=\frac{1}{\frac{\left(\sum \limits_{i=1}^{n} \beta_{i}^{2}\right) \cdot\left(\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}\right)}{\left(\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}^{-1}\right)}}=\frac{1}{\bar{\lambda}\left(\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}^{-1}\right)} \\\end{array}$$
For the second part: If I now compare the F(x) with the inequality $\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}^{-1} \leq \frac{\lambda_{1}+\lambda_{n}-\lambda}{\lambda_{1} \lambda_{n}}$ results in the following: $$ F(x) \geq \frac{1}{\bar{\lambda} \cdot \frac{\lambda_{1}+\lambda_{n}-\lambda}{\lambda_{1} \lambda_{n}}}=\frac{\lambda_{1} \lambda_{n}}{\lambda_{1}+\lambda_{n}-\bar{\lambda}} $$ $\bar{\lambda}$ as the mean value of the eigenvalues $\lambda_{i}$: $$ \bar{\lambda}=\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}=\sum \limits_{i=1}^{n}\left(\sum \limits_{j=1}^{n} \beta_{j}^{2}\right)^{-1} \beta_{i}^{2} \lambda_{i}=\frac{\sum \limits_{i=1}^{n} \lambda_{i}}{\sum \limits_{j=1}^{n} \beta_{j}^{2}} $$
This results in $$ F(x) \geq \frac{\lambda_{1} \lambda_{n}}{\lambda_{1}+\lambda_{n}-\frac{\lambda_{i=1}^{n} \lambda_{i}}{\sum \limits_{j=1}^{n} \beta_{j}^{2}}}=\frac{\sum \limits_{j=1}^{n} \beta_{j}^{2}}{\left(\lambda_{1}+\lambda_{n}\right) \sum \limits_{j=1}^{n} \beta_{j}^{2}-\sum \limits_{i=1}^{n} \lambda_{i}}=\frac{4 \lambda_{1} \lambda_{n}}{\left(\lambda_{1}+\lambda_{n}\right)^{2}}=4 \lambda_{1} \lambda_{n}\left(\lambda_{1}+\lambda_{n}\right)^{-2} $$
Is this then a lower bound of the function $\mathrm{F}(\mathrm{x})$?