Square root proof question

[melkorthefoul]

1. The problem statement, all variables and given/known data

Hi guys,

I'm doing a rather long math modelling task. As part of the task, I have to investigate the following argument:

There is a rectangle ABCD. It is sliced by the line EF, such that EF is parallel to AD and BC. M is the midpoint of BC. X is a point on the line EF (the position of X, which is denoted as x, is the variable in this investigation).

If, for a given value of x, the square of the distance AX is minimized, then for this value of x the distance AX is also minimized (proved that). Then, if, for a given value of x, the square of the distance XM is minimized, then for this value of x the distance XM is also minimized (Proved that the same way). Therefore, if the sum of the squares of the distances AX and XM are minimized for a given value of x, then the sum of the distances (AX+XM) is also minimized (Need just a bit of advice here)

2. Relevant equations

AX2=f(x)
XM2=g(x)
AX2+XM2=h(x)
AX+XM=i(x)
3. The attempt at a solution

I've kinda proved the last bit as well, by saying that the square root of (a+b) does not equal the square root of a plus the square root of b, as the argument assumes that the square root of h(x) = i(x) . However, I just wanted to clarify something: would saying just this be enough, or do I have to prove it? And how would I go about doing that?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[Mark44]

1. The problem statement, all variables and given/known data

Hi guys,

I'm doing a rather long math modelling task. As part of the task, I have to investigate the following argument:

There is a rectangle ABCD. It is sliced by the line EF, such that EF is parallel to AD and BC. M is the midpoint of BC. X is a point on the line EF (the position of X, which is denoted as x, is the variable in this investigation)

If, for a given value of x, AX2 is minimized, then for this value of x AX is also minimized (proved that). Then, if, for a given value of x, XM2 is minimized, then for this value of x XM is also minimized (Proved that the same way). Therefore, if (AX2+XM2) is minimized for a given value of x, then (AX+XM) is also minimized (Need just a bit of advice here)

2. Relevant equations

AX2=f(x)
XM2=g(x)
AX2+XM2=h(x)
AX+XM=i(x)
This is all very confusing. A, B, C, D, X and M are the names of points. What are AX and XM supposed to represent? Is AX the line segment between A and X? Is AX2 the square of the magnitude of AX?

3. The attempt at a solution

I've kinda proved the last bit as well, by saying that the square root of (a+b) does not equal the square root of a plus the square root of b, as the argument assumes that the square root of h(x) = i(x) . However, I just wanted to clarify something: would saying just this be enough, or do I have to prove it? And how would I go about doing that?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[vela]

This is all very confusing. A, B, C, D, X and M are the names of points. What are AX and XM supposed to represent? Is AX the line segment between A and X? Is AX2 the square of the magnitude of AX?
Yeah, I found it confusing too, but it's a convention used in high-school geometry. The line segment between two points A and X is denoted \overline{AX} whereas its length is denoted AX, without the overline.

[melkorthefoul]

Whoops.... clarified what AX and XM stand for. My bad :D

[Mark44]

The minimum value of AX + XM is attained when the point X is on the line segment AM. In that case, AX + XM = AM. If X is not on the line AM, AX + MX > AM.