inverse triangle inequality in $L^p$ I requiere some inequalities generalization. I don't know whether they are true or not. Can you help me ?
Here we're talking about $L^p$ spaces with $p > 1$.

I know that on the real line :
$$ ||x|-|y|| \leq | x-y | \leq |x|+|y| $$ equivalently : 
$$ ||x|-|y|| \leq | x+y | \leq |x|+|y|$$

Now i'm trying to find similar inequalities in Lebesgues spaces.
I already found that one :
$$(|x + y|)^p \leq 2^{p-1} (|x|^p + |y|^p)$$ thanks to Jensen ineqality.
I also know Minkowski inequality's telling me :
$$ \|f + g\|_{L^p} \leq \|f\|_{L^p} + \|g\|_{L^p}$$

Now I'm searching for something on the other boundary. Meaning, as my
  friends told me should be true :
$$ |\|f\|_{L^p} - \|g\|_{L^p} | \leq  \|f-g\|_{L^p}$$ and
  equivalently :
$$ |\|f\|_{L^p} - \|g\|_{L^p} | \leq  \|f+g\|_{L^p}$$

i would also like to find something like this :
$$\lambda |(|x|^p - |y|^p)| \leq (|x + y|)^p $$ 
Do you know if something like those 2 inequalities exists, and if yes, how do you prove them ?
thanks !
 A: In any normed space $$||x||=||(x-y)+y||\le||x-y||+||y||,$$hence $$||x||-||y||\le||x-y||.$$Similarly (or "hence, swapping $x$ and $y$"), $$||y||-||x||\le||x-y||.$$ If $a,b\in\Bbb R$ then $|a-b|=\max(a-b,b-a)$, so 
$$\big|\,||x||-||y||\,\big|\le||x-y||.$$
A: The result in the quote box is true by exactly the same proof as in $\mathbb R$. 
The second part of your question is essentially a duplicate of this older question of mine, Can I expect $|x|^s - |y|^s \leq C|x-y|^s$ for $s>1$? (with related discussion) where a counterexample (valid already in dimension 1) shows that $||x|^p - |y|^p| \le C |x-y|^p$ is not possible for $p>1$. in 1D, the counterexample is obtained by choosing $$x=x_0,y=x_0 + t, \quad x_0 > 0,\quad  0<t\ll 1.$$
Essentially, this is because $|x|^p$ is $C^1$ when $p>1$ so we have the (local) estimate from Mean Value Theorem,
$$  |x+h|^p - |x|^p  \approx   p\operatorname{sgn}(x)|x|^{p-1} h.$$
That is, the leading error term is linear in $h$, and its not possible to obtain a bound with a higher order error term than $|h|$.
