About some inequalities on integrals Well, I've got the next exercise.
Prove the if $f$ and $g$ are integrable over [a,b] and for $p>1$ then: $$\int_{a}^{b} |f(x) + g(x)|^p dx \leq 2^p\int_{a}^{b}|f(x)|^p dx + 2^p\int_{a}^{b} |g(x)|^p dx$$
Proof (attempt): For the triangle's inequality we've got that $$|f(x) + g(x)| \leq |f(x)| + |g(x)|$$ Thus, suppose WGL that $|g(x)|= \max{(|f(x)|,|g(x)|)}$ Thus: $$ |g(x)|>|f(x)| \implies |g(x)|^p > |f(x)|^p $$ 
At this is when I'm stocked, the intention is to prove that $| f + g|^p \leq |f|^p + |g|^p$ and the idea of the last line is very useful according to my professor, but I don't see how. If someone can give me any hint or a different way to get through the proof. I'd be thankful.
 A: If you are still confused for the proof. Let me help you a bit.
Define $\phi(y) = |y|^{p}$ for $p>1$. Thus, we know that $\phi$ is a convex function for any $y \in \mathbb{R}$. Observe the following Jensen's inequality for convex function ( you can find the proof easily on internet) :
$$ \phi(ty_{1} + (1-t)y_{2}) \leq t\phi(y_{1}) + (1-t)\phi(y_{2}) \quad \forall t \in \mathbb{R}, \, 0<t<1 $$
Therefore, by taking $y_{1} = f(x)$ and $y_{2} = g(x)$ and setting $t=\frac{1}{2}$, we have the following result:
\begin{equation}
\begin{aligned}
 |\frac{1}{2}f(x)+\frac{1}{2}g(x)|^{p} &\leq \frac{1}{2}|f(x)|^{p} + \frac{1}{2}|g(x)|^{p} \\
2^{-p}|f(x)+g(x)|^{p} &\leq 2^{-1}\big( |f(x)|^{p} + |g(x)|^{p}\big)\\
|f(x)+g(x)|^{p} &\leq 2^{p-1}\big( |f(x)|^{p} + |g(x)|^{p}\big)
\end{aligned}
\end{equation}
Therefore, we conclude that $ |f(x)+g(x)|^{p} \leq 2^{p-1}\big( |f(x)|^{p} + |g(x)|^{p}\big) \leq 2^{p}\big( |f(x)|^{p} + |g(x)|^{p}\big) $.
By using the monotonicity of integral operator, you will get your desired result.
