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 $\geq$