Max norm of product normes spaces Edit: I need some help with this. 
Let $$(V_1, \|·\|_1)$$ and $$(V_2, \|·\|_2)$$ be normed spaces, and the product space $$V = V_1\times V_2$$
be endowed with the norm
$$\|(x_1, x_2)\| = \max\{ \|x_1\|_1, \|x_2\|_2 \}.$$
Show that if $V_1$ and $V_2$ are Banach spaces then $$(V, \|(·, ·)\|)$$ is a a Banach space.
I tried to prove that $$\|(x_1, x_2)\| = \max\{ \|x_1\|_1, \|x_2\|_2 \}.$$ is a norm but I am stuck with showing triangle inequality. Do I need to use the fact that $$\|x\|_1 = \sum_{i=0}|x_i|$$ and that $$\|x\|_2 = \sqrt{\sum_{i=0}|x_i|^2}$$ or it is not necessary?
I had problems with showing completeness (which means that I need to find a convergent Cauchy sequence) but it was answered in the reply below. Thanks for your help.
 A: Let $(v_n,w_n)_{n\in\Bbb N}\subset V$ be a cauchy sequence and $\epsilon >0$. Then there is $N$ such that $$\|(v_N,w_N)-(v_{N+n},w_{N+n})\|< \epsilon\qquad \forall n\geq 0$$. In particular $$\|v_{N}-v_{N+n}\|_1\leq \|(v_N,w_N)-(v_{N+n},w_{N+n})\|<\epsilon\qquad \forall n\geq 0$$ i.e. $(v_n)_{n\in\Bbb N}\subset V_1$ is a Cauchy sequence. By a similar argument we know that $(w_n)_{n\in\Bbb N}\subset V_2$ is a Cauchy sequence. Thus there are $v,w$ such that $$\lim_{n\to \infty}v_n = v\qquad \text{ and }\qquad \lim_{n\to \infty}w_n=w.$$
In particular there are $N_1,N_2$ such that $\|v_n-v\|_1<\epsilon$ for every $n > N_1$ and $\|w_n-w\|_1<\epsilon$ for every $n > N_2$. It follows that
$\|(v_n,w_n)-(v,w)\|<\epsilon$ for every $n>\max\{N_1,N_2\}$.
EDIT: In order to show that $\|\cdot\|$ is a norm, you can use the fact that $\|\cdot \|_1$ and $\|\cdot\|_2$ are norms. E.g. for the triangle inequality we get for every $(v,w),(x,y)\in V$ that
$$\|v+x\|_1\leq \|v\|_1+\|x\|_1\leq\|(v,w)\|+\|(x,y)\|$$
and
$$\|w+y\|_2\leq \|w\|_2+\|y\|_2\leq\|(v,w)\|+\|(x,y)\|$$
thus $$\|(v,w)+(x,y)\| \leq\|(v,w)\|+\|(x,y)\|$$.
