Upper-bound $\sup_{x \ne 0}\dfrac{\sum_i \lambda_i x_i^2}{\sum_i \sigma_i x_i^2}$, where $\lambda_1 \ge \ldots \ge \lambda_n > 0$ (same with $\sigma$)

Question

Let $\lambda_1 \ge \ldots \ge \lambda_n > 0$ and $\sigma_1 \ge \ldots \ge \sigma_n > 0$ and for nonzero $x=(x_1,\ldots,x_n) \in \mathbb R^n$, define the quotient $$ R(x) := \dfrac{\sum_{i=1}^n \lambda_i x_i^2}{\sum_{i=1}^n \sigma_i x_i^2}. $$

Question. What is a good upper-bound for $R^* := \sup_{x \in \mathbb R^n,\;x \ne 0}R(x)$ ?

Trivial bound. It is easy to see that $R^* \le \lambda_1/\sigma_n$, but this bound is really bad, as can be seen when $\lambda_i=\sigma_i$ for all $i$. Indeed, this trivial bound predicts $R^* \le \lambda_1/\lambda_d$ which can be arbitrarily larger than the actual value of $1$.

Question 1. What if $\lambda_i = \sigma_i$ for all $i \ge k$ (where $k \le n$ is fixed), is there a good upper-bound for $R^*$ as a function of $k$ and the $\lambda$'s and $\sigma$'s ?

Answer 1

By scaling the vector $x$, we may assume $$\sum_{i=1}^n \sigma_ix_i^2=1,$$ and we want to maximize $$\sum_{i=1}^n \lambda_ix_i^2.$$ Substituting $y_i=\sqrt{\sigma_i}x_i$, we have $\sum y_i^2=1$ and want to optimize $$\sum_{i=1}^n \left(\frac{\lambda_i}{\sigma_i}\right)y_i^2.$$ This is at most $$\max_{1\leq i\leq n}\frac{\lambda_i}{\sigma_i}\sum_{i=1}^n y_i^2=\max_{1\leq i\leq n}\frac{\lambda_i}{\sigma_i},$$ and equality is reached by one of the elementary vectors ($1$ in one component and $0$ in all others). So, the optimum is $\max \lambda_i/\sigma_i$.

Answer 2

You can use $$\min_i \frac{a_i}{b_i} \le \frac{\sum_i a_i}{\sum_i b_i}\le \max_i \frac{a_i}{b_i}$$ ( $b_i$ positive)