I'm trying to show the function $f: \mathbb{R}^2 \to \mathbb{R}$ defined by $f(v) = (\sqrt{|v_1|} + \sqrt{|v_2|}$ is not a norm on $\mathbb{R}^2$ by showing the triangle inequality doesn't hold. That is, I want to show $\Vert u + v \Vert > \Vert u \Vert + \Vert v \Vert$.
I believe I have correctly simplified the problem to a comparison between $\sqrt{|u_1| + |v_1|}\sqrt{|u_2| + |v_2|}$ on the LHS and $\sqrt{|u_1||u_2|} + \sqrt{|v_1||v_2|}$ on the RHS. I don't see how I can say with certainty that $\sqrt{|u_1| + |v_1|}\sqrt{|u_2| + |v_2|} > \sqrt{|u_1||u_2|} + \sqrt{|v_1||v_2|}$.
Am I missing some operation on square roots I can do or some other simplification or maybe a substitution? Perhaps I'm barking up the right tree altogether and should show $\Vert\alpha v\Vert \neq |\alpha|\Vert v\Vert$ instead?