Absolutely absolute

To minimize the value, it's clear we must minimize | x + 1 |, as all of the other values are positive constants. For any real y, | y | is always non-negative. So, in our problem, the only value of x that yields minimum | x + 1 | is x = -1, and the expression reduces to 2+3+4 = 9

First of all we have to analyze the main absolute value function: $f(x)=|x+1|$ As we can see it can be considered has:

$f(x)=x+1$ if $x\geq-1$

$f(x)=-x-1$ if $x<-1$

That got a zero in the point $A: (-1;0)$

We know that an absolute value function is, obviously, always positive, and we only have a sum with positive integers. The sum of two positive numbers will always give a positive number... So we can get rid of all the others absolute value functions we got. $g(x)= \large \Bigg | \bigg | \Big | \big | x + 1 \big | + 2 \Big | + 3 \bigg | + 4 \Bigg | = |x+1|+2+3+4 = |x+1| +9$

So, that funcion is just a translation of f(x) upwards (by 9 units). All things considered we can say that the point of minimum has the same x-value of the point A:

$g(-1)=|-1+1| +9=9$ The minimum value is $\boxed{9}$

Bonus: If we consider a generic function of the form:

$h(x)= \large \Bigg | \bigg | \Big | \big | x + a \big | + b \Big | + c \bigg | + d \Bigg |$

If $b; c; d$ are non-negative values. The maximum point is : $M(-a;k)$ with $k=b+c+d$

That is true for all absolute values nested multiple times if the numbers we add to the central absolute value function are non-negative.