Solve: $\|u+v\| \le \|u\| + \|v\|$ with $\|x\| = \left( \sqrt{|x_1|} + \sqrt{|x_2|} \right)^2$ I was given the following task:

Check if $x\rightarrow \left(\sqrt{|x_1|} + \sqrt{|x_2|}\right)^2$ is a norm on $\mathbb{R}^2$.

I've already shown that 
$$\|x\| \ge 0\qquad \|x\| = 0 \Leftrightarrow x = 0$$
$$\|\alpha x\| = |\alpha| \cdot \|x\|$$
The last thing that I need to show is the triangle inequality using $\sqrt{|a|+|b|}\le \sqrt{|a|} + \sqrt{|b|}$
$$\|u+v\| = \left(\sqrt{|u_1+v_1|} + \sqrt{|u_2+v_2|}\right)^2$$
$$\le \left(\sqrt{|u_1|} + \sqrt{|v_1|}+ \sqrt{|u_2|} + \sqrt{|v_2|} \right) ^2$$
$$= \left(\sqrt{|u_1|} + \sqrt{|u_2|}+ \sqrt{|v_1|} + \sqrt{|v_2|}\right)^2$$
$$\dots$$
I have no idea how to continue at this point. I've also tried a different approach:
$$\|u+v\| = \left(\sqrt{|u_1+v_1|} + \sqrt{|u_2+v_2|}\right)^2$$
$$\ge {\sqrt{|u_1|+|u_2|+|v_1|+|v_2|}}^2 = |u_1|+|u_2|+|v_1|+|v_2|$$ 
but this didn't get me any further as well.
I am thankful for any help

Let $u(a^2,b^2)$ and $v(c^2,d^2),$ where $a$, $b$, $c$ and $d$ are positives.
Thus, we need to prove that
$$(a+b)^2+(c+d)^2\geq\left(\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\right)^2$$ or
$$ab+cd\geq\sqrt{(a^2+c^2)(b^2+d^2)},$$ which can be wrong by C-S.
