# Quadratic mean inequality where arguments are linearly combination with 1

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.

## 1 Answer

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 $\lambda$.