Does $|x|^p$ with $0I am curious about whether $|x|^p$ with $0<p<1$ satisfy $|x+y|^p\leq|x|^p+|y|^p$ for $x,y\in\mathbb{R}$. 
So far my trials show that this seems to be right... So can anybody confirm whether this is right or wrong and give a proof or counterexample?
 A: By homogeneity, we try to see whether $|1+t|^p\leq 1+|t|^p$ for all $t$, where $0&ltp&lt1$. We first look whether it's the case if $t\geq 0$. Define the maps $f(t):=(1+t)^p-t^p-1$ on $\mathbb R_{\geq 0}$. The derivative is $f'(t)=p((1+t)^{p-1}-t^{p-1})\leq 0$, so $f(t)\leq f(0)=0$ and the inequality holds for $t\geq 0$. To deal with the general case, note that 
$$|1+t|^p\leq |1+|t||^p\leq 1+||t||^p=1+|t|^p.$$
This inequality is useful to deal with the $L^p$ spaces for $0&ltp&lt1$, showing that the map $d(f,g)=\int |f-g|^pd\mu$ is a metric.
A: Since $\frac{d^2 x^p}{dx^2} = p (p-1) x^{p-2} &lt 0$, the function $x^p$ is a convex function for $x>0$, therefore for $x,y >0$ the inequality is a special case of Jensen's inequality.
$$
  \frac{x^p+y^p}{2} \geq \left(\frac{x+y}{2}\right)^p =\frac{2}{2^p} \frac{(x+y)^p}{2} \geq \frac{(x+y)^p}{2}
$$
For negative $x$ or $y$ one has $|x| + |y| \geq |x + y|$. Therefore 
$$
\frac{|x|^p+|y|^p}{2} \geq \frac{(|x|+|y|)^p}{2} \geq  \frac{|x+y|^p}{2}.
$$
(Well, there are already two answers)
A: This is a special case of (the very general) inequality (2.12.2) from theorem (27) in Hardy, Littlewood, Polya's "Inequalities". 
(As this is homework I just tell you it's true and let you have a look at a classic if you really want to cheat with the proof. -- Ups, Davide already provided a proof... :-)
A: Notice that
\begin{equation}
1. \quad p-1 < 0 \Rightarrow t^{p-1} >(t+|x|)^{p-1} \quad \forall t \in (0,\infty).
\end{equation}
Then, integrate 1 from  0 to $|y|$ to obtain the desired inequality.
