Why the triangle inequality is greater than the 2 max{f(x),g(x)}

[cbarker1]

TL;DR Summary

I am working on proving the L^p of measureable space (X,S,u) is a vector space. I am lost on why the the triangle intequalty is greater than the 2 max{f(x),g(x)} for a fix x in X.

I am reading Sheldon's Axler Book on Measure theory. He is proving that $L^p(\mu)$ is a Vector space over $\mathbb{R}.$ He claims that if $\|f+g\|_{p}^p\leq 2^p(\|f\|_{p}^p+\|g\|_{p}^{p})$ and nonzero homogenity holds true, then $L^p{\mu}$ is true with the standard addition and scalar multiplication. He starts with the following assumption:

Suppose that $f,g\in L^p(\mu)$ are arbitrary. Then if $x\in X$ is an arbitrary fix element of $X,$ then
$\begin{align*} |f(x)+g(x)|^p&\leq_{\text{triangle inequality}} (|f(x)|+|g(x)|)^p\\ &\leq_{\text{why?}} (2\max{|f(x)|,|g(x)|})^p\\ &\leq2^p(|f(x)|^p+|g(x)|^p)\end{align*}$

If you can explain whys in this proof then I will be able to understand the proof.

Thanks,

Carter Barker

 

[Hill]

Because if, e.g., |f|>=|g|, then |f|+|g|<=|f|+|f|=2|f|