Isometry of $(V\times W)^*$ with $V^*\times W^*$

I need to prove that $J:V^*\times W^*\to (V\times W)^*$ is a isometry, with $J(f,g)(v,w)=f(v)+g(w)$. Where $V$ and $W$ are normed spaces, further $\|(v,w)\|_{V\times W}=\max\{\|v\|_V,\|w\|_W\}$ and $\|(f,g)\|_{(V\times W)^*}=\|f\|_{V^*}+\|g\|_{W^*}$.

So far by a direct calculation I have managed to show that $$\|J(f+g)\|=\sup_{\|(v,w)\|=1}|f(v)+g(w)|\leq\sup_{\|(v,w)\|=1}(\|f\|_{V^*}+\|g\|_{W^*})\cdot \|(v,w)\|=\|(f,g)\|_{(V\times W)^*}$$

But the other inequality did not come out until now, but I will tell you what I came up with, taking $w=0$, we have $J(f,g)(v,0)=f(v)$, then $\|J(f,g)\|\geq \|f\|$ but I get nowhere.

On the other hand it also occurred to me to take a particular $(v_0,w_0)$ pair such that $|J(f,g)(v_0,w_0)|\geq\|f\|_{V^*}+\|g\|_{W^*}$ but it seems to be a silly idea. Any better ideas?

Fix $\varepsilon>0$. There exist $v_0\in V$, $w_0\in W$, with $\|v_0\|=\|w_0\|=1$ and $$ |f(v_0)|+\varepsilon >\|f\|,\qquad |g(w_0)|+\varepsilon >\|g\|. $$ By multiplying each by an appropriate number in the circle, we may assume that $$ f(v_0)+\varepsilon >\|f\|,\qquad g(w_0)+\varepsilon >\|g\|. $$ Then, since $\|v_0,w_0\|=1$, $$ \|J(f,g)\|\geq J(f,g)(v_0,w_0)=f(v_0)+g(w_0)>\|f\|+\|g\|-2\varepsilon. $$ As this occurs for all $\varepsilon>0$, $$ \|J(f,g)\|\geq\|f\|+\|g\|=\|(f,g)\|. $$