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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 $\geq$

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    $\begingroup$ A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = \|\sqrt{M}x\|^2$. Thus, we can rewrite your equation as $$ \frac{\|A^{1/4}x\|^2}{\|A^{1/2}x\|} < \frac{\|A^{1/2}x\|^2}{\|Ax\|} \iff\\ \frac{\|Ax\|}{\|A^{1/2}x\|} < \frac{\|A^{1/2}x\|^2}{\|A^{1/4}x\|^2} $$ with $B = A^{1/4}$ and $y = A^{1/4}y$, we can rewrite the above as $$ \frac{\|B^3y\|}{\|By\|} < \frac{\|By\|^2}{\|y\|^2} \iff \|B^3y\|\,\,\|y\|^2 < \|By\|^3 $$ $\endgroup$
    – Ben Grossmann
    Commented Apr 23, 2019 at 1:44
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    $\begingroup$ Also, note that we fail to have strict inequality when $A = I$, for instance. $\endgroup$
    – Ben Grossmann
    Commented Apr 23, 2019 at 2:06
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    $\begingroup$ More thoughts that are insufficient for an answer: Since both sides scale with $\|y\|$, it suffices to consider the inequality in the case that $\|y\| = 1$. That is: $$ \|B^3y\| \leq \|By\|^3 $$ To that end: consider $$ \min \|By\|^6 - \|B^3y\|^2 \quad \text{st} \quad \|y\|=1 $$ Let $f(y) = \|By\|^6 - \|B^3y\|^2$, and let $g(y) = \|y\|^2$. We compute the Lagrangian $$ 2B^2(3\|By\|^4 I - \lambda B^4)y $$ Now, $B$ positive definite. So, setting the Lagrangian to zero yields $$ (3\|By\|^4 I - \lambda B^4)y = 0 \implies \left(\frac{3 \|By\|^4}{\lambda} I - B^4\right)y = 0 $$ $\endgroup$
    – Ben Grossmann
    Commented Apr 23, 2019 at 2:59
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    $\begingroup$ Applying the Hölder-von Neumann inequality yields $$ \|By\|^2 = \operatorname{tr}[B^2yy^T] \leq \operatorname{tr}[B^3]^{1/1.5}\operatorname{tr}[(yy^T)^{3}]^{1/3} = \operatorname{tr}[B^3]^{2/3}\|y\|^{2/3} $$ which is close to what we're looking for, but not quite there $\endgroup$
    – Ben Grossmann
    Commented Apr 23, 2019 at 3:05

1 Answer 1

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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.

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