Help with inequality for real numbers: $||x+y|^t-|x|^t-|y|^t|\leq C(|x|^{t-1}|y|+|x||y|^{t-1})$ I have a problem with proving the following inequality
$$||x+y|^t-|x|^t-|y|^t|\leq C(|x|^{t-1}|y|+|x||y|^{t-1})$$
where $t>1$ and $x,y$ are real numbers and $C$ is only dependent on $t$.
I tried to use the Hölder inequality $|a|^s|b|^{1-s}\leq s|a|+(1-s)|b|$, triangle inequality and about any other trick I know of, but have been getting nowhere.
The closest I got is
$$
LHS\leq (2^{t-1}+1)(|x|^t+|y|^t)+2^{t-1}(|x|^{t-1}|y|+|x||y|^{t-1})
$$
but I don't think I can improve it to be $\leq RHS$.
 A: I am not sure that this is the most straightforward approach, or that there is any way of using known inequalities to get the result, but at least it seems to give a proof of the stated inequality.
Wlog we can assume that both $x$ and $y$ are different from zero. Indeed, if not, the stated result reduces to the elementary inequality $0 \leqslant 0$, which is obviously true. Start by distinguishing two cases: either both $x$ and $y$ have the same sign or not. Since $t > 1$, it is easy to see that
$$
|x+y|^t - |x|^t - |y|^t \geqslant 0
$$
holds in the first case, while the opposite inequality is true if $xy < 0$.
Case I ($\mathbf{xy > 0})$: In this case we want to show
$$
|x+y|^t-|x|^t-|y|^t \leqslant  C(|x|^{t-1}|y|+|x||y|^{t-1}).
$$
By symmetry we can further assume that $x \leqslant y$. Dividing both terms by $|y|^{t}$ and setting $z = \frac{x}{y}$ we are left to prove the following inequality
$$
|z+1|^t - |z|^t  - 1 \leqslant C ( |z|^{t-1} + |z| )
$$
under the constraint $0 \leqslant z \leqslant 1$. Basic calculus then gives that there exists a constant $C$ such that
$$
(1+z)^t \leqslant 1 + C z + C z^{t-1} + z^t
$$
holds. Indeed, the function $(1+z)^t$ is continuous with bounded derivative for $z \in [0,1]$ and $t >1$. Moreover, for $z = 0$ identity holds in the above relation. Therefore choosing $C$ sufficiently large we deduce
$$
(1+z)^t \leqslant 1 + C z
$$
for $z \in [0,1]$ which implies the desired estimate.
Case II ($\mathbf{xy < 0})$: Wlog we can assume $|x| \geqslant |y|$ and $x > 0$. Arguing as in the previous case, the desired inequality can be reduced to
$$
|z|^t + 1 \leqslant |z-1|^t + C|z|^{t-1} + C|z|
$$
for $z \in [0,1]$. Again this follows easily since the function $|z|^t + 1$ is continuous with bounded derivative and identity holds for $z = 0$.
P.S.: The first case in the argument above seems to give $C = 2^t-1$ as best possible constant. However, this does not appear to be optimal in general.
