How to show that $(a+b)^p\le 2^p (a^p+b^p)$ If I may ask, how can we derive that $$(a+b)^p\le 2^p (a^p+b^p)$$ where $a,b,p\ge 0$ is an integer? 

Assume that both $A, B > 0$ and $p \geq 0$ to begin with, then we can rewrite the inequality as:
$\left(\dfrac{A}{A+B}\right)^p + \left(\dfrac{B}{A+B}\right)^p \geq 2^{-p}$. So there are $3$ cases to consider:


*

*$p = 0$, then $LHS = 2 > 1 = RHS$, and the inequality holds.

*$p = 1$, then $LHS = 1 > \dfrac{1}{2} = RHS$, and the inequality also holds.

*$p \geq 2$, then $f(x) = x^p$ is convex, so $f(x) + f(y) \geq f\left(\dfrac{x+y}{2}\right)$
with $x = \dfrac{A}{A+B}$, and $y = \dfrac{B}{A+B}$ yields the inequality.
