Upper estimate for positive definite matrix Suppose $\Lambda=\text{diag}(\lambda_1,\dots,\lambda_d)$ is $d\times d$ diagonal matrix with all diagonal entries being positive. For fixed $\alpha\in \mathbb{R^d}$ with $||\alpha||^2=d$, I want to get upper estimate for
\begin{aligned}
\frac{(\alpha^t\Lambda \alpha)^{\frac{r}{2}}}{\alpha^t\Lambda^{\frac{r}{2}}\alpha}
\end{aligned}
,where $r$ is fixed positive integer, depending only on $d$ and $r$. If $r=1$ or $2$, one can easily observe that the upper bound is given by $1$. But if $r\geq 3$, to obtain estimate is quite hard. In case $r\geq 3$, my attempt was as following:
$$(\alpha^t\Lambda\alpha)^{\frac{r}{2}}=(\sum_{i=1}^d\lambda_i\alpha_i^2)^{\frac{r}{2}}\leq(\sum_{i=1}^d \lambda_i)^{\frac{r}{2}}(\sum_{i=1}^d \alpha_i^2)^{\frac{r}{2}}\leq d^{\frac{r}{2}-1}(\sum_{i=1}^d \lambda_i^{\frac{r}{2}})(\sum_{i=1}^d\alpha_i^2)^{\frac{r}{2}}=d^{r-1}(\sum_{i=1}^d \lambda_i^{\frac{r}{2}})$$.
But since the lower estimate of the denominator $\alpha^t\Lambda^{\frac{r}{2}}\alpha$ is difficult to obtain, I'm getting trouble in finding the desired upper estimate for $\frac{(\alpha^t\Lambda \alpha)^{\frac{r}{2}}}{\alpha^t\Lambda^{\frac{r}{2}}\alpha}$. Can anyone help? Thanks in advance.
 A: Welcome to MathStackExchange! I think Holder's inequality (https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality) works here.
The target function $\frac{(\alpha^t\Lambda\alpha)^\frac{r}{2}}{\alpha^t\Lambda^{\frac{r}{2}}\alpha} = \frac{(\sum\limits_{i=1}^n \lambda_i \alpha_i^2)^{\frac{r}{2}}}{\sum\limits_{i=1}^n\lambda_i^{\frac{r}{2}}\alpha_i^2}$.
To be more concise, please allow me to define $\beta_i = \alpha_i^2 \geq 0$ and $s = \frac{r}{2} \geq \frac{3}{2}$. Then the target function becomes $\frac{(\sum\limits_{i=1}^n \lambda_i \beta_i)^{s}}{\sum\limits_{i=1}^n\lambda_i^s\beta_i}$.
The Holder inequality states:
If $x,\,y$ be two vectors in $\mathbb{R}^n$ and $p,\,q$ are two real numbers in ($1$, $+\infty$), then $\sum\limits_{i=1}^n{|x_iy_i|} \leq (\sum\limits_{i=1}^n|x_i|^p)^{\frac{1}{p}}(\sum\limits_{i=1}^n|x_i|^q)^{\frac{1}{q}}$ if $\frac{1}{p} + \frac{1}{q} = 1$. The critical step is to find $x_i, y_i, p, q$. That is not difficult, we may set:
$x_i = \lambda_i\beta_i^\frac{1}{s}$, $y_i = \beta_i^{\frac{s-1}{s}}$, $p=s$, $q=\frac{s}{s-1}$.
Following the Holder inequality, $\sum\limits_{i=1}^n \lambda_i\beta_i \leq (\sum\limits_{i=1}^n \lambda_i^s\beta_i)^{\frac{1}{s}}(\sum\limits_{i=1}^n \beta_i)^\frac{s-1}{s}$, therefore the answer to the question is $(\sum\limits_{i=1}^n \beta_i)^{s-1} = d^{s-1}$.
