R^3 Triangle inequality norm

Question

Let's define norm in $\mathbb{R}^3$ $$ ‖x‖ = \sqrt{x_1 ^2 + 2x_2 ^2 + 3x_3 ^2}$$ How can I solve triangle inequality $||x + y|| \leq ||x|| + ||y||$. I did: $$||x+y|| = \sqrt{(x_1 + y_1)^2 + 2(x_2 + y_2)^2 + 3(x_3 + y_3)^2}$$ and I don't know how to get $$\sqrt{x_1 ^2 + 2x_2 ^2 + 3x_3 ^2} + \sqrt{y_1 ^2 + 2y_2 ^2 + 3y_3 ^2}$$ Can you help me solve this? Or maybe this is seminorm?

Answer 1

As @Anne Bauval suggested to you, $\|x+y\|=\|f(x+y)\|_2=\|f(x)+f(y)\|_2\leq\|f(x)\|_2+\|f(y)\|_2$ because $\|.\|_2$ is a norm.

And I'll let you conclude and possibly add an argument that I would have implied.