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? 

In fact we can make the stronger claim:
$$(a+b)^p\le 2^{p-1} (a^p+b^p)$$
if $a,b \ge 0$. As Mohammad Khosravi points out, this is equivalent to
$$(x+1)^p\le 2^{p-1} (x^p+1)$$
if $x \ge 0$. We prove this by induction. The case $p=1$ is easy: $x+1 \le x+1$. So suppose we have established that
$$(x+1)^{p-1}\le 2^{p-2} (x^{p-1}+1)$$
Then we have:
$$\begin{align}
(x+1)^p & = (x+1)(x+1)^{p-1}\\
& \le 2^{p-2}(x+1)(x^{p-1}+1)
\end{align}$$
Now we just have to show that $2^{p-2}(x+1)(x^{p-1}+1) \le 2^{p-1} (x^p+1)$, or equivalently that $2(x^p+1) - (x+1)(x^{p-1}+1) \ge 0 $:
$$\begin{align}
2(x^p+1) - (x+1)(x^{p-1}+1) & = x^p-x^{p-1}-x+1 \\
& = (x-1)(x^{p-1}-1) \\
& = (x-1)^2(x^{p-2} + x^{p-3} + \ldots + x + 1) \\
& \ge 0
\end{align}$$
