TITLE: Proving inequality for positive definite matrix
QUESTION [6 upvotes]: For a positive definite diagonal matrix $A$, I want to prove that for any $x$:
$$\frac{x^T \sqrt{A} x}{\|\sqrt{A}x\|_2} \geq \frac{x^T A x}{\|Ax\|_2}$$
So far I cannot find any counterexamples, and it intuitively makes sense since the $\sqrt{\cdot}$ operator should bring the eigenvalues of $A$ closer to $1$, but I can't prove this.

EDIT: changed $>$ to $\geq$

REPLY [14 votes]: Your inequality says
$$\frac{\sum\sqrt{\lambda_j}x_j^2}{\left(\sum\lambda_j x_j^2\right)^{1/2}}\geq 
\frac{\sum\lambda_jx_j^2}{\left(\sum\lambda_j^2x_j^2\right)^{1/2}},$$
or after a simple transformation
$$\sum\lambda_j x_j^2\leq\left(\sum\sqrt{\lambda_j}x_j^2\right)^{2/3}
\left(\sum\lambda_j^2x_j^2\right)^{1/3}$$ 
And this is Holder's inequality with 
$p=3/2$ and $q=3$. The strict inequality does not always hold.