@Kitonum 

It is really diffficult to rank the many solutions provided here, but yours is brilliant while being more versatile

The Sylvester's criterion is often use to check if square real valued matrix can be a variance matrix : as I am concerned with this issue your procedure will be of great help here too.

Thanks for all the work you did

@acer 
You write "FWIW"  ...
In french  "For What It's Worth" has some kind of negative significance, something like "I give you this information but it's not worth much, just look if you can find any interest in it".
Maybe its english significance is subtly different ?
For, personally, I find it very instructive (so from "french acceptation" this information matters), specially the one contained in the first grey rectangle : thinking to split the problem (more generally a problem) is an idea that could be useful in other situations.


Thanks acer
 

@John May 
I had used the command solve this way
solve(r^2-r*s+s^2 = 0, r)
and next checked if the solutions were complex (wich gave just a part of the answer) ... but it looked far too articifial

The way you handle the issue is undoubtedly more clever than mine.

Great thanks

@vv  it's all clear to me now.

You wrote :  is(I, positive)  is of course false (not being real) ... I now understand why

Thanks a lot

@vv  I have just done those few elementary tests : all if them return the correct "true" answer

restart;

assume(r, positive):

assume(s, positive):           

is(r^2+s^2, positive);                                  

is(r^2+s^2 > 0);                   

restart:                               

is(r^2+s^2 > 0) assuming r > 0, s > 0;                   

is(r^2+s^2 > 0) assuming positive                      

@vv ...  unfortunately disappointing answer (Maple handles only relatively simple conditions when dealing with several variables)

Point [1]
Strangely I did not obtained a FAIL answer by using
is( r^2 - r*s + s^2> 0) assuming r >0, s>0;

or
assume(r>0): #or assume(r, positive)
assume(s>0): #or assume(s, positive)
is( r^2 - r*s + s^2> 0) ;

but always "false"

Note : even writing r^2 - r*s + s^2  in the following form   1/2*(r - s)^2 + (1/2)*(r^2  + s^2 ) has the same result (which suggests checking r^2+s^2 will also fail (?)).


Point [2]
I guess you refer to my auxiliary message  ?
            is(I, 'complex')                          # obviously true
            is(I, 'positive')                          # returns false
I know the set of real numbers is contained in the set of real numbers and, as a rule Maple operates over the latter , but how can Maple decide if a pure imaginary number is positive ? Or some complex number ?

Finally I did know the way you use assuming in the commang
is( r^2 - r*s + s^2> 0) assuming positive;
What are the inderterminates "assuming positive" refers to ?

Thanks again for the time spent

assume(r, 'positive')
is(r, 'complex')                 # returns  true, just as coulditbe does too
                                        # Is this suggest that being 'positive' is consistent with being complex ?

Now let's try this
is(I, 'complex')                          # obviously true
is(I, 'positive')                           # returns false

then an object of complex type is not positive (at first hand I would have thought Maple will return FAIL ???)
 

@Markiyan Hirnyk I will remember that

@Kitonum ... I should have find this by myself !

Nevertheless thank vou for reminding me my dumbness :-)

@tomleslie  ... and I'm aware of that.
While maybe too radical, your suggestion  remains interesting.

Thanks

@John May  Nice way to proceed.

@tomleslie  Even if not complete I do appreciate your contribution

@acer   Great thanks to you for this very documented answer.

Concerning your last question, the answer is yes.
The worksheet is very small but it launches a whole computational code whose procedures print useful intermediate informations.
Once executed the worksheet mainly contains output results (possiblyredirected to a text file through the writeto command); using "printf"  thus enables a more readable "execution report".

@Markiyan Hirnyk 

In order to be prefectly clear, should I have mention that before using SetEdgeWeight I had used MakeWeighted ?

# G = some unweighted graph
WG := MakeWeighted(G)   # gives weights 1 to all edges in G
# link = some edge in WG
SetEdgeWeight(WG, link, infinity) # does not work

Is this sufficiently accurate for you ?

I thank you again for your first answer, all the rest is a waste of time
 

@Markiyan Hirnyk 

As I had already constructed an unweighted graph (not from its adjacency matrix), I found using the procedure  SetEdgeWeight was a simple and fast way go introduce weights.
But it does accept only numeric values as written in the Maple 2015' help pages (I did not try with Maple 2016)

It is unfortunate that MakeWeighted and SetEdgeWeight do not accept the same objects ...

Nevertheless I thank you for your useful solution