# norm of product of normed spaces

If $(X_1,||.||_1)$ and $(X_2,||.||_2)$ are two normed spaces and define norm on $X_1\times X_2$ as $||x||=\max(||x_1||_1,||x_2||_2)$. I want to check the triangle inequality property for this norm, i.e, $||x+y||\leq||x||+||y||$. After using using triangle inequality for the two norms I get $||x+y||\leq\max(||x_1||_1+||y_1||_1,||x_2||_2+||y_2||_2)$. My question is why is $\max(||x_1||_1+||y_1||_1,||x_2||_2+||y_2||_2)\leq\max(||x_1||_1,||x_2||_2)+\max(||y_1||_1,||y_2||_2)$

$$\lVert x_1\rVert_1+\lVert y_1\rVert_1\le \max\{\lVert x_1\rVert_1,\lVert x_2\rVert_2\}+\max\{\lVert y_1\rVert_1,\lVert y_2\rVert_2\}$$ $$\lVert x_2\rVert_2+\lVert y_2\rVert_2\le \max\{\lVert x_1\rVert_1,\lVert x_2\rVert_2\}+\max\{\lVert y_1\rVert_1,\lVert y_2\rVert_2\}$$
• Thanks, I had to do the 4 cases of $x$ and $y$ and verified this result. May 1 '13 at 14:57