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.




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EDIT: changed $>$ to $\geq$



Proving inequality for positive definite matrix
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](https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality) with 
$p=3/2$ and $q=3$. The strict inequality does not always hold. 


