Showing the inequality $|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)$ I have a small question that I think is very basic but I am unsure how to tackle since my background in computing inequalities is embarrassingly weak - 
I would like to show that, for a real number $p \geq 1$ and complex numbers $\alpha, \beta$, I have
\begin{equation}
|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)
\end{equation}
I thought it would be best to rewrite this as 
\begin{equation}
\left|\frac{\alpha + \beta}{2}\right|^p \leq \frac{|\alpha|^p + |\beta|^p}{2}
\end{equation}
but then I am unsure what to do next - is this a sensible start anyways ? Any help would be great !
(P.S. this is not a homework question - I am currently trying to brush up my knowledge of $L^p$ spaces, and this inequality came up as a statement. I thought it might be worthwhile to  make sure I can fill in the gaps to improve my skills in computing inequalities.)
 A: Without loss of generality, we can assume that that $\lvert \alpha \rvert \geq \lvert \beta \rvert$. Now essentially you want to prove that $$\displaystyle \left \lVert \frac{z + 1}{2} \right \Vert^p \leq \displaystyle \frac{\lVert z \rVert^p + 1}{2},$$ where $\displaystyle z = \frac{\alpha}{\beta}$ and $\lVert z \rVert \geq 1$.
Note that $$ \displaystyle \left \lVert \frac{z + 1}{2} \right \Vert \leq \frac{\lVert z \rVert + 1}{2}$$
So if we prove that $\displaystyle \left(\frac{1+t}{2} \right)^p \leq \frac{1+t^p}{2}$, where $t = \lVert z \rVert \geq 1$, we are done.
Now this is a one-variable calculus problem. Consider $f(t) = \displaystyle \frac{1+t^p}{2} - \left(\frac{1+t}{2} \right)^p$. We then get that $$\displaystyle f'(t) = \frac{p t^{p-1}}{2} - \frac{p (1+t)^{p-1}}{2^p}.$$ Hence, $$f'(t) = \frac{p}{2} \left( t^{p-1} - \left(\frac{1+t}{2} \right)^{p-1}\right) \geq 0$$ for $t \geq 1$. Hence, $f(t)$ is increasing for all $t \geq 1$. And $f(1) = 0$. Hence, we have that $$f(t) \geq f(1) = 0.$$ Hence, we get that $$\displaystyle \left(\frac{1+t}{2} \right)^p \leq \frac{1+t^p}{2},$$ where $t = \lVert z \rVert \geq 1$
A: In fact, one advantage of the argument in john w.'s answer is that it can be extended further  to prove $$(a+b)^n\leq p^na^n+q^nb^n,$$ where $\frac{1}{p}+\frac{1}{q}=1$. In particular, one can choose the coefficient of $a^n$ very close to $1$ by paying the price of a larger coefficient of $b^n$. 
The proof is achieved by arguing either $(a+b)\leq pa$ or $(a+b)\leq qb$ holds. This is true since otherwise one would get a contradiction that $\left(\frac{1}{p}+\frac{1}{q}\right)(a+b)>a+b$.
A: If the $p-1$ being $p$ is not a big deal then I think the following works.
$|\alpha+\beta|\leq |\alpha|+|\beta|\leq 2\max\{|\alpha|,|\beta|\}$.  If the max is $|\alpha|$, then $|\alpha+\beta|^p\leq 2^p|\alpha|^p\leq 2^p(|\alpha|^p+|\beta|^p)$. Similarly for $\beta$. In either case you get the inequality you want.
A: *

*The map $x\mapsto x^p$ for $x\geq 0$ is convex, since its second derivative is $p(p-1)x^{p-2}\geq 0$. 

*We have 
$$\left|\frac{a+b}2\right|^p\leq \left(\frac{|a|+|b|}2\right)^p\leq \frac{|a|^p+|b|^p}2.$$

A: To generalize this $L_p$ (p>1) norm for $\alpha$ and $\beta$, we can introduce weights such that $q+r=1$, so we have
\begin{align}
q\alpha+r\beta=(q\alpha^p)^{1/p}q^{1-1/p}+(r\beta^p)^{1/p}r^{1-1/p}\underset{Holder}{\le}\left(q\alpha^p+r\beta^p\right)^{1/p}(q+r)^{1-1/p}=\left(q\alpha^p+r\beta^p\right)^{1/p}
\end{align}
therefore,
\begin{align}
(q\alpha+r\beta)^p\le q\alpha^p+r\beta^p.
\end{align}
In case of $q=r=\frac12$, the originally asked is then obtained.
\begin{align}
\left|\frac{\alpha + \beta}{2}\right|^p \leq \frac{|\alpha|^p + |\beta|^p}{2}
\end{align}
Note the sign "$\le$" becomes "$\ge$" for the case $0<p<1$.
