$\sigma(x,y) = |x_{1} - y_{1}|^3 + |x_{2} - y_{2}| + ... + |x_{n} - y_{n}|$, how would I show that this is equivalent but not uniformly equivalent? If $x,y \in \mathbb{R}^n$, and $\sigma(x,y) = |x_{1} - y_{1}|^3 + |x_{2} - y_{2}| + ... + |x_{n} - y_{n}|$, how would I show that this is equivalent but not uniformly equivalent? I already figured out that this is a metric, but I am having a hard time showing as I stated above. Clearly the power of $3$ is not uniformly continuous which means its not uniformly equivalent. Just to clear any confusion I am looking for a set of $n$ tuples that is equivalent but uniformly equivalent which is why I chose the following metric above. The $\epsilon - \delta$ argument for equivalence is not as easy as I thought it would be. The help would be appreciated!

Hint
The metrics are uniformly equivalent if and only if there exist $c,C>0$ such that 
$$c(|x_{1} - y_{1}| + |x_{2} - y_{2}| + \cdots + |x_{n} - y_{n}|)\le \sigma(x,y)\le C( |x_{1} - y_{1}| + |x_{2} - y_{2}| + \cdots + |x_{n} - y_{n}|).$$
In particular, for $x=(x_1,x_2,\cdots,x_n),y=(y_1,x_2,\cdots,x_n)$ it should be
$$c |x_{1} - y_{1}| \le |x_{1} - y_{1}|^3 \le C|x_{1} - y_{1}|,$$ or equivalently,
$$c \le |x_{1} - y_{1}|^2 \le C.$$
Is this possible?
