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 pairwise orthonormal eigenvectors. For $x \neq 0$ let $$ F(x):=\frac{\left(x^{\top} x\right)^{2}}{\left(x^{\top} A x\right)\left(x^{\top} A^{-1} x\right)} $$
Show $$ F(x)=\frac{1}{\bar{\lambda}\left(\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}^{-1}\right)} $$ $\operatorname{with} x=\sum \limits_{i=1}^{n} \beta_{i} u_{i}, \beta_{i} \in \mathbb{R}, \gamma_{i}:=\left(\sum \limits_{j=1}^{n} \beta_{j}^{2}\right)^{-1} \beta_{i}^{2}, i=1, \ldots, n $, and $\bar{\lambda}:=\sum \limits_{i=1}^{n} \gamma_{i} \lambda_{i}$.
I'm not quite sure how to show this task - $\left(x^{\top} x\right)$ can I express it as follows $$ x^{\top} x=\left(\sum \limits_{i=1}^{n} \beta_{i} u_{i}\right)^{\top}\left(\sum \limits_{j=1}^{n} \beta_{j} u_{j}\right)=\sum \limits_{i=1}^{n} \sum \limits_{j=1}^{n} \beta_{i} \beta_{j} u_{i}^{\top} u_{j}=\sum \limits_{i=1}^{n} \beta_{i}^{2} $$ and $\left(x^{\top} A x\right)$ and $\left(x^{\top} A^{-1} x\right)$ as $$ \begin{array}{c} x^{\top} A x=\sum \limits_{i=1}^{n} \beta_{i}^{2} \lambda_{i} \\ x^{\top} A^{-1} x=\sum \limits_{i=1}^{n} \beta_{i}^{2} \lambda_{i}^{-1} \end{array} $$
If this is correct, do I just have to substitute these results into F(x)?
Yes, your expressions for $x^\top x, x^\top Ax, x^\top A^{-1}x$ are all correct. From there, one nice way to derive the desired formula is as follows: \begin{align} \frac 1{F(x)} & = \frac{(x^\top Ax)(x^\top A^{-1}x)}{(x^\top x)^2} = \frac{x^\top Ax}{x^\top x} \cdot \frac{x^\top A^{-1}x}{x^\top x} \\ & = \frac{\sum_{i=1}^n \beta_i^2\lambda_i}{\sum_{i=1}^n \beta_i^2} \cdot \frac{\sum_{i=1}^n \beta_i^2\lambda_i^{-1}}{\sum_{i=1}^n \beta_i^2} \\ & = \frac{\sum_{i=1}^n \beta_i^2\lambda_i}{\sum_{j=1}^n \beta_j^2} \cdot \frac{\sum_{i=1}^n \beta_i^2\lambda_i^{-1}}{\sum_{j=1}^n \beta_j^2} \\ & = \left(\sum_{i=1}^n \frac{\beta_i^2}{\sum_{j=1}^n \beta_j^2} \lambda_i\right) \cdot \left(\sum_{i=1}^n \frac{\beta_i^2}{\sum_{j=1}^n \beta_j^2} \lambda_i^{-1}\right) \\ & = \left(\sum_{i=1}^n \gamma_i \lambda_i\right) \cdot \left(\sum_{i=1}^n \gamma_i \lambda_i^{-1}\right) = \bar \lambda \left(\sum_{i=1}^n \gamma_i \lambda_i^{-1}\right). \end{align} Taking the reciprocal on both sides yields the desired formula.