Inequality between the norm of derivative and the derivative of norm

Let $x(t)=[x_1(t)~x_2(t)~\cdots ~x_n(t)]^T$, function $x_i:R\rightarrow R$ is differentiable, then it can be drawn that when $p=2$, $\|\frac{d}{dt}x(t)\|_p\geq \frac{d}{dt}\|x(t)\|_p$. I wonder if this inequality holds when $p\neq 2$, and if it does, show why.

• Please give more context... Is it $L^p(\mathbb R)$ or $L^p(\mathbb \Omega)$ for a particular $\Omega$ ? also what is $x$ ? do you mean $x(t)\in L^p(\Omega)$ for every $t$ ? and if yes what are the assumptions on $x$ ? – Renart Feb 25 '16 at 12:49
• As it is stated, this inequality is always true: The left hand side is something wich is nonnegative and the right hand side is the derivative of a constant, hence zero. If you mean by the $\|\cdot\|_p$ norm, the $p$-norm for vectors in $\mathbb R^n$ and if $x:\mathbb R\to\mathbb R^n$ is a differentiable curve then your question gets a meaning. Is this what you mean? – frog Feb 25 '16 at 12:57

$$\left|\frac{\|x(t+h)\|-\|x(t)\|}{h}\right|≤\left\|\frac{x(t+h)-x(t)}{h}\right\|$$
From this consideration you get that whenever $\partial_t x(t)$ and $\partial_t \|x(t)\|$ exist in a normed vector space, then you must have
$$\partial_t \|x(t)\|≤\big|\partial_t\|x(t)\|\big|≤\|\partial_t x(t)\|$$