Let $(\mathcal{X}, \mathcal{A}, \mu)$ be a measure space. Take $f,g \in L^1(\mu)$. Prove that $\sqrt{f^2+g^2}$ and $\sqrt{\vert fg\vert}$ are in $L^1(\mu)$.
First, I prove that $h = \max\{f,g\} \in L^1(\mu)$. Let $A = \{x \in \mathcal{X} \mid f(x) \geq g(x)\}$. Then:
$$ \begin{align} \int_{\mathcal{X}} h \ \mathrm{d}\mu &= \int_A h \ \mathrm{d}\mu + \int_{A^c} h \ \mathrm{d}\mu \\ \int_{\mathcal{X}} h \ \mathrm{d}\mu &= \int_Af \ \mathrm{d}\mu + \int_{A^c} g \ \mathrm{d}\mu \\ \int_{\mathcal{X}} h \ \mathrm{d}\mu &\leq \int_{\mathcal{X}}f \ \mathrm{d}\mu + \int_{\mathcal{X}} g \ \mathrm{d}\mu < +\infty \end{align}$$
So $h \in L^1(\mu)$ as I wanted. If I suppose that $f,g \geq 0$, I think the following will do: let's prove that $\sqrt{f^2 + g^2} \in L^1(\mu)$. We have:
$$\begin{align} \int_{\mathcal{X}} \sqrt{f^2 + g^2} \ \mathrm{d}\mu &\leq \int_{\mathcal{X}} \sqrt{2\max\{f,g\}^2} \ \mathrm{d}\mu = \sqrt{2}\int_{\mathcal{X}} \vert\max\{f,g\}\vert \ \mathrm{d}\mu < \infty\end{align}$$
since $\max\{f,g\} \in L^1(\mu) \iff |\max\{f,g\}| \in L^1(\mu)$. I'm a bit insecure about this.
And for the other one:
$$\int_{\mathcal{X}} \sqrt{\vert fg\vert} \ \mathrm{d}\mu \leq \int_{\mathcal{X}} \sqrt{|\max\{f,g\}^2|}\ \mathrm{d}\mu = \int_{\mathcal{X}} \max\{f,g\}\ \mathrm{d}\mu < + \infty.$$
Not secure about this one, either.
Can someone help me fixing the details? Alternate solutions are also welcome. Thank you.
Ashamed of myself for not realizing the problem was so elementary, I'll register my solution here, in case it helps someone. We have that for all $a,b \in \Bbb R$: $$\sqrt{a^2 + b^2} \leq \sqrt{|a|^2 + 2|ab| + |b|^2} = \sqrt{(|a|+|b|)^2} = ||a|+|b|| = |a|+|b|$$
and: $$(|a|-|b|)^2 \geq 0 \implies a^2 - 2|ab| + b² \geq 0 \implies |ab| \leq 2|ab| \leq a^2 + b^2 \implies \sqrt{|ab|} \leq \sqrt{a^2 + b^2}.$$
Applying this to the functions, we obtain:
$$\int_{\mathcal{X}} \sqrt{f^2 + g^2} \ \mathrm{d}\mu \leq \int_{\mathcal{X}} |f| + |g| \ \mathrm{d}\mu = \int_{\mathcal{X}} |f| \ \mathrm{d}\mu + \int_{\mathcal{X}} |g| \ \mathrm{d}\mu < +\infty$$
since $f,g \in L^1(\mu) \implies |f|,|g| \in L^1(\mu)$. We conclude that $\sqrt{f^2+g^2} \in L^1(\mu)$. And now, we just use that: $$\int_{\mathcal{X}} \sqrt{|fg|} \ \mathrm{d}\mu \leq \int_{\mathcal{X}} \sqrt{f^2 + g^2} \ \mathrm{d}\mu < +\infty$$ and hence $\sqrt{|fg|} \in L¹(\mu)$.