Lower bound for $\vert x-y\vert^p$ Let $x,y\in\mathbb{R}$ and $1<p.$
Is is possible to get a constant $K_{p}$ such that $$\vert x-y\vert^p\ge K_{p}\left(\vert x\vert^p\pm\vert y\vert^p\right)?$$
I tried using the function $f(t)=t^p$ or convexity but I don't get a result.
I ask because it happens some time we have a increasing function $g$ and we want an "linear" upper bound for $g(-\vert x-y\vert^p)$ of course the $-$ is interesting for integrability for exemple.
 A: The bound $|x-y|^p\geq K_p(|x|^p+|y|^p)$ cannot hold for some $K_p>0$ and for all $x,y\in\mathbb R$: setting $x=y=1$ for example, that would show that $K_p\leq 0$, a contradiction.
Also, the bound $|x-y|^p\geq K_p(|x|^p-|y|^p)$ cannot hold: this would imply that, for $y=1$ and for all $x>1$, $$K_p(x^p-1)\leq (x-1)^p\,\,\Rightarrow\,\,K_p\frac{x^p-1}{x-1}\leq (x-1)^{p-1}.$$ Then, since $p>1$, the right hand side of the last inequality goes to $0$ as $x\to 1^+$, while the left hand side goes to $pK_p$, therefore $K_p\leq 0$, a contradiction.
Edit: the inequality $|x-y|^p\geq K_p|x|^p+L_p|y|^p$ for $K_p,L_p>0$ still cannot hold, by taking $x=y=1$.
On the other hand, if $\displaystyle f(t)=\frac{(t+1)^p}{t^p+1}$ for $t>0$, then $$f'(t)=\frac{p(t+1)^{p-1}(t^p+1)-(t+1)^ppt^{p-1}}{(t^p+1)^2}=\frac{p(1-t^{p-1})}{(t^p+1)^{p+1}},$$ hence $f$ achieves its maximum when $t=1$. Therefore, for every $t>0$, $(t+1)^p\leq 2^{p-1}(t^p+1)$, so for every $y\neq 0$, $$\left(\frac{|x-y|}{|y|}+1\right)^p\leq 2^{p-1}\left(\frac{|x-y|^p}{|y|^p}+1\right)\,\,\Rightarrow \left(|x-y|+|y|\right)^p\leq 2^{p-1}\left(|x-y|^p+|y|^p\right).$$ The last inequality shows that, if $y\neq 0$, $|x|^p\leq 2^{p-1}|x-y|^p+2^{p-1}|y|^p$, therefore $$|x-y|^p\geq 2^{1-p}|x|^p-|y|^p.$$ The last inequality also holds for $y=0$ and any $x$, so it holds for any $x,y\in\mathbb R$.
