Let $p \geq 4$ be an even integer. In the study of variational problems in $W^{1, p}$, it is handy to know that for $a, b \in \mathbb R^d$,
$$|a - b|^p \leq 2^{p - 1} (|a|^{p - 2} + |b|^{p - 2}) |a - b|^2 \leq 2^{p - 2} \langle |b|^{p - 2} b - |a|^{p - 2} a, b - a\rangle.$$
Indeed this inequality implies that the Comparison Principle holds for the $p$-Laplacian.

My question is if the following matrix analogue holds. Let us use the Frobenius inner product on square matrices. Let $Q(A) := \sqrt{AA^\dagger}$ denote the positive-semidefinite part of $A \in \mathbb R^{d \times c}$.

> Does there exist $C = C(d, c) > 0$ such that for every $A, B \in \mathbb R^{d \times c}$,
$$|A - B|^p \leq C^p \langle Q(B)^{p - 2} B - Q(A)^{p - 2} A, B - A\rangle?$$

Such an inequality would be useful to prove certain estimates on the [Daskalopoulous--Uhlenbeck Schatten $p$-Laplacian][1].

I drew $1000$ matrices in $\mathbb R^{2 \times 2}$ with entries in $[-1, 1]$ at random and checked this inequality for $4 \leq p \leq 20$ on them. It seems to hold with $C < 1.6$. By scale-invariance, the boundedness of the entries should not be very important.

To try to prove this inequality, let's first observe that the second derivative of $A \mapsto \operatorname{tr}(Q(A)^p)$ in the direction $A - B$ is the sum over all traces $\operatorname{tr}(C_1 C_2^\dagger C_3 \cdots C_{p - 1} C_p^\dagger)$ where $p - 2$ of the $C_j$s are equal to $A$, and the remaining $2$ are equal to $A - B$. Let $S_A$ be the set of all $p$-tuples $\vec C$ with this property (and similarly $S_B$). Meanwhile the first derivative is $p\langle Q(A)^{p - 2} A, A - B\rangle$. I don't think the third derivatives are particularly important. If we discard all terms in the Taylor expansion which involve the third derivative and higher we "conclude" that 
$$\langle Q(B)^{p - 2} B - Q(A)^{p - 2} A, B - A\rangle \geq \frac{1}{pC^p} \sum_{\vec C \in S_A} \operatorname{tr}(C_1 \cdots C_p^\dagger) + \frac{1}{pC^p} \sum_{\vec C \in S_B} \operatorname{tr}(C_1 \cdots C_p^\dagger).$$
So we "just" need to prove
$$|A - B|^p \leq C^p  \sum_{\vec C \in S_A} \operatorname{tr}(C_1 \cdots C_p^\dagger) + C^p \sum_{\vec C \in S_B} \operatorname{tr}(C_1 \cdots C_p^\dagger).$$
(I am resorting to the usual analyst convention of letting $C$ be a different constant in each line.)
The Frobenius norm is rather unnatural for this problem and it's better to work with the $p$-Schatten norm, 
$$|A - B|^p \leq C^p \operatorname{tr}(Q(A - B)^p) = C^p \operatorname{tr}(((A - B)(A - B)^\dagger)^{p/2}).$$
Expanding out this $p/2$th power we get a sum over traces $\operatorname{tr}(C_1 \cdots C_p^\dagger)$ where $p - 2$ of the $C_i$s are either $A$ or $-B$ and the last two are $A - B$. If we could commute these matrices then we would complete the proof (this is basically the proof in the scalar case, anyways). But these matrices do not commute (and if $d \neq c$ it doesn't even make sense to commute them). So this approach might be a dead end.


  [1]: https://arxiv.org/abs/2205.08250