To show $|f(x) + g(x)| \le 2 \max(|f(x)|, |g(x)|)$ implies $|f(x) + g(x)|^2 \le 4 (|f(x)|^2 + |g(x)|^2)$

Question

In the proof of showing $L^2(\mathbb{R}^d)$ vector space, with functions $f$ and $g$ both belonging to $L^2(\mathbb{R}^d)$ ...

We know $|f(x) + g(x)| \le 2 \max(|f(x)|, |g(x)|)$.

Squaring both sides, we get

$|f(x) + g(x)|^2 \le 4 \max(|f(x)|, |g(x)|)^2$.

From this, how does Stein and Shakarchi (vol 3, pg 158) get the sum from the max: $|f(x) + g(x)|^2 \le 4 (|f(x)|^2 + |g(x)|^2)$ ?

Answer 1

Consider the domain to be $\Bbb R^{+}=$ set of positive reals .

For any $x;$ $x= \sqrt {x^2}\le\sqrt {x^2+y^2}$

$\implies \max\{x,y\}\le \sqrt {x^2+y^2}$ .