0
$\begingroup$

Is it true that for any $\alpha \in [0, 1]$, and numbers $\lambda_i$ for $i = 1, 2, ..., n$, such that $\sum_{i=1}^n \lambda_i = n$ and $\forall_i \lambda_i \geq 0$ the following inequality holds?

$$ \sqrt{\sum_{i=1}^n \frac{1}{\lambda_i^2}} \geq \sqrt{\sum_{i=1}^n \frac{1}{\left(\alpha \lambda_i + 1-\alpha \right)^2}} $$

For $\alpha = 1$ it holds trivially. For $\alpha = 0$ it holds from inequality between quadratic mean and harmonic mean. Some low dimensional cases have checked by hand seem to work. Can't see how to prove / disprove it. Maybe some smart application of Jensen's?

Thank you for any pointers / suggestions.

$\endgroup$
1
$\begingroup$

The point defined by $\mu_i = \alpha\lambda_i + (1-\alpha)$ is any point on the segment between $\lambda$ and $(1,\ldots,1) \in \mathbb{R}^n$. This segment is included in the subspace $\sum_i x_i = n$.

Now consider the function $$f\colon(x_1,\ldots,x_{n-1})\mapsto\sum_{i=1}^{n-1} \frac{1}{x_i^2} + \frac{1}{(n-x_1-\ldots-x_{n-1})^2}. $$ This function is clearly convex and its first derivative is $$ \frac{\partial f}{\partial x_i} = \frac{-2}{x_i^3} + \frac{2}{(n-x_1-\ldots-x_{n-1})^3}. $$ We conclude that $(1,\ldots,1) \in \mathbb{R}^{n-1}$ is the minimum. So that it is increasing when it goes from $(1,\ldots,1) \in \mathbb{R}^n$ to $\lambda$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.