@Markiyan Hirnyk 

If in Mathematica one wants to retrive Kitonum's solutions, a similar by hand substitution will be needed.

@Markiyan Hirnyk 

You are right. I have corrected the second example, which was answered correctly by Maple.

@Markiyan Hirnyk 

Workarounds? But these are bugs too!

@tomleslie 

Yes, it is Pi^2/8 with "probability 1".

It reduces to
J:=2*Int(arcsinh(cosh(x))-x, x=0..infinity);
identify(evalf(J));  #Pi^2/8

is(value(J)=Pi^2/8); #false

It seems to be hard to compute J symbolically.

@John Fredsted 

Theoretically it could be formulated as sequence of optimization problems:
find the vectors vi = [xi1,xi2,...,xi8] (i=1,...,n) satisfying the constraints ||vi||=1, ||vi-vj|| >= 1.

Then increase n until a solution does not exist.

But practically the number of the constraints n*(n+1)/2 is probably too large for n>50 say, to hope for a solution.

@John Fredsted 

Why do you say that? A double-check is always better than none!

Edit. I mean e.g. to use Maple to find a maximal set of vectors V on the sphere ||x||=r in R^8 such that ||x-y|| >= r, for x,y in V, x<>y.

@John Fredsted 

A simple inspection is enough. E.g. if both x,y have integer entries then x-y can have 1,2,3 or 4 nonzero entries, and in the case of a single nonzero entry, this must be +/-2. So ||x-y|| >= sqrt(2) [attained if x-y has 2 nonzero entries +/-1].

P.S. It would be nice to use Maple to find such a configuration.

I don't know much about sphere packing, but I don't understand why do you need a Maple check here, when mathematically this is almost obvious: for E8, if x,y are distinct "root" vectors, ||x||=sqrt(2), ||x-y|| >= sqrt(2).
Probably in one of the Leech constructions the situation is similar.

Optimization:-Maximize uses floats (just like DS), so, the true max could be 2 + exp(-10^99999999999).

@acer 

It seems that Or considers (here) only two arguments.

Conclusion: it's hard to live in the Maple world!

@acer 

OK, thank you, so

simplify(combine(expand(r^2))) assuming Or(x<0,x=0,x>0), Or(y<0,y=0,y>0);

works, and my infinity prediction was "highly exagerated".
I like the important ingredient Or(x<0,x=0,x>0).

Edit. I though it was a typo, but
simplify(combine(expand(r^2))) assuming Or(x<0,x=0,x>0), Or(y<0,y=0,y>0);
simplify(combine(expand(r^2))) assuming Or(x<0,x=0,x>0), Or(y=0,y<0);
indeed do not simplify in a fresh session, but do simplify later (due to remember tables)!

@Carl Love 

I can't either.
And I suspect that the number of necessary commands to simplify is infinity. :-)

(Actually, using polar coordinates x= a*cos(t), y=a*sin(t) it can be done with assumptions on t, but it's not elegant.)

Note also the strange answers:

is(r^2=x+I*y) assuming real;  # FAIL
is(r^2=x+I*y) assuming positive; # false

@Carl Love 

If

r:=evalc(sqrt(x+I*y));

how many commands are needed to simplify r^2 to x+I*y,  assuming x,y real?

You keep saying "incorrect" etc, when all my statements are ok.

cp returns (as I said) a single expression (with RootOf) valid for both min and max. It is your allvalues which gives some of the multiple values (see the _ValuesMayBeLost value which is set to true). Some are complex as you see.
The user must do the following:

1. Find the critical points (contained in the last parameter of extrema).
2. Select the real ones c1,....
3. Compute f(c1),...
4. Compute their max and min

The difficulty is that (as I said) there are an infinity of critical points. Fortunately in this example it is easy to see that only three are real.

I hope that everything is clear for you now.

[And please, stop suspecting me of not knowing maths; as I said to you, I am a professional mathematician with prizes  at IMO (long time ago), many published papers,  etc].

@Markiyan Hirnyk 

Sorry, I cannot see what exactly does not correspont to reality.

@Markiyan Hirnyk