Bound on triangle inequality with powers

Question

Conjecture

I believe the following inequality is true but I cannot prove it. Let $\alpha \in (0, 2)$ and $x, y \in \mathbb{R}$. Then, it holds that

$$ \Big\vert \vert x + y \vert^\alpha - \vert x \vert^\alpha - \vert y \vert^\alpha \Big\vert \leq \begin{cases} 2 \vert x \vert^{p\alpha} \vert y \vert^{(1 - p)\alpha} &, 0 < \alpha \leq 1 \\ \alpha \vert x \vert \vert y \vert^{\alpha - 1} + (\alpha + 1) \vert x \vert^{p\alpha} \vert y \vert^{(1 - p)\alpha} &, 1 < \alpha \leq 2 \end{cases} $$ for any $p \in [0, 1]$.

Edit

The inequality follows surprisingly easily from symmetry of $x$ and $y$ in Equation 3.8 of Astrauskas et al (1991) as suggested by Brendan McKay. This inequality reads as $$ \Big\vert \vert x + y \vert^\alpha - \vert x \vert^\alpha - \vert y \vert^\alpha \Big\vert \leq \begin{cases} 2 \vert x \vert^{\alpha} &, 0 < \alpha \leq 1 \\ \alpha \vert x \vert \vert y \vert^{\alpha - 1} + (\alpha + 1) \vert x \vert^{\alpha} &, 1 < \alpha \leq 2 \end{cases} $$

In that paper, the inequality was stated as a matter of fact without any reference. I would prefer it if I could cite an original source with proof instead of just the equation from Astrauskas et al (1991). Therefore, I have added a reference-request tag to this question.

Answer 1

In this answer we will focus on the inequalities listed in Astrauskas et al (1991).

First observe that if $y = 0$ then the inequality is trivial. So we can assume $y \neq 0$. Then replacing $y \mapsto 1$ and $x\mapsto \hat{x} = x/ y$, the scaling homogeneity of the problem shows it suffices to prove this case; that is, to bound the quantity $$ \left||x+1|^\alpha - |x|^\alpha - 1\right| $$

This can be done as a calculus problem. I'll do the first case $(\alpha \in (0,1])$ below. I believe the second case can be done similarly.

Your goal is to find the maximum value of $$ A(x) = \frac{ ||x+1|^\alpha - |x|^\alpha - 1|}{|x|^\alpha} $$

When $x \geq 0$, we have $A(x) = 1 + \frac{1 - |x+1|^\alpha}{|x|^\alpha}$. Taking derivatives you find $A'(x) \leq 0$ with $A'(x) = 0\iff x = 0$. At the origin L'Hopital's rule gives $A(0) = 1$.

When $x \leq -1$, a similar argument finds $A'(x) > 0$.

On the interval $(-1,-0)$, a computation shows that $A'(x) < 0$. And hence the maximum is attained at $x = -1$ where $A(-1) = 2$.