Show that $(\int_\Omega |\nabla u|^2 + \int_\Omega |u|^2)^{1/2}$ is a norm in $H_0^1$

Question

I am reviewing some material on Sobolev spaces and taking some time to derive the basics again. But I am unable to see how to show the triangle inequality for

$$ ||u|| = \left( \int_\Omega|\nabla u|^2 dx + \int_\Omega|u|^2\right)^{1/2}. $$

I tried using the triangle inequality for the $L^2$ norm, but to no end:

$$ ||u+v|| = (||\nabla u + \nabla v||_2^2 + ||u + v||_2^2)^{1/2} \\ \leq \left( (||\nabla u||_2 + ||\nabla v||_2)^2 + (||u||_2 + ||v||_2)^2 \right)^{1/2} \\ \leq ||\nabla u||_2 +||\nabla v||_2 +||u||_2 + ||v||_2 $$

Any hint will be the most appreciated.

Thanks in advance.

Answer 1

As the comment by @user58955 suggested, use the Minkowski inequality (this process works for $W^{k,p}$):

\begin{align*} \|u+v\|&=\left(\|\nabla u+\nabla v\|_2^2+\|u+v\|^2\right)^{1/2}\\ &\leq \left(\left(\|\nabla u\|_2+\|\nabla v\|_2\right)^2+\left(\|u\|_2+\|v\|_2\right)^2\right)^{1/2}\\ &\leq \left(\|\nabla u\|_2^2+\|u\|_2^2\right)^{1/2}+\left(\|\nabla v\|_2^2+\|v\|_2^2\right)^{1/2}\\ &=\|u\|+\|v\|. \end{align*} This is how it is done, for example, in Evans.