Why does minimizing absolute value and squares of residuals in a regression give different answers? We are minimizing either the $1$ norm of the residuals, least absolute value, or the $2$ norm, least squares. 
Least absolute value:
$$\min_\beta||y - x \beta||_1$$
Least squares:
$$\min_\beta||y - x \beta||_2$$
But by the equivalence of norms, they only differ by some constants $c$, $C$:
$$c||y - x \beta||_1 \le ||y - x \beta||_2 \le C||y - x \beta||_1$$
If they are only different by a constant, why then is the minimizer different?
 A: They aren't different by a constant. If that were so, you would be right. But what the formula you post says is that they are different by at least some constant and at most some other constant.  Further, the norms are the single numbers that are the total for all the values. But the individual contributions of each point can vary.
For the least squares, the contribution of extreme points is higher than for absolute value. 
A: Let $x$ be scalar and equal to one to consider the most simple example. Then consider minimizing
$$\min_\beta \mathbb E[(Y-\beta)^2]$$
The first order condition:
$$0 = -2 \int_{-\infty}^\infty (Y - \beta)dF(y) = -2(\mathbb E[Y] - \beta)$$
implying that the solution for minimizing $\beta$ is $\beta^* = \mathbb E[Y]$. 
Consider then the problem $$\min_\beta \mathbb E[\lvert Y - \beta\lvert]$$
again differentiating (absolute norm is convex so we can use subgradients from convexity theory to generalize the differential operator) to get the first order condition
$$0 = \int_{-\infty}^\infty \delta(Y-\beta) dF(y)$$
where $\delta(.)$ is the subgradient of the absolute norm and hence $\delta(Y-\beta) = - 1$ for $Y<\beta$ and $\delta(Y-\beta) =  1$ when $Y>0$. Therefore
$$0 = \int_{-\infty}^\infty \delta(Y-\beta) dF(y) = - \int_{-\infty}^\beta dF(y) + \int_{\beta}^\infty dF(y)$$
which is equivalent to
$$0 =  - F(\beta)+  (1-F(\beta))$$
implying that
$$F(\beta) = \frac{1}{2}$$ hence the solution for $\beta$ must be $\beta^* = F^{-1}(1/2)$ which is the median of the distribution $F$.
Using sample moments instead of expectations the solution become respectively the sample mean and the sample median. And in case of covariates $x$ it result in standard regression in the case of squared errors and median regression in case of the absolute norm.
