@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

@Markiyan Hirnyk 

I know, but I do not see anything wrong in my previous comment. On the contrary, the information should be useful for anyone using extrema or similar commands.

@Markiyan Hirnyk 

In this example the situation is the same. extrema returns the values (min and max) for the functions at the critical points (but complex included). In this case a single expression but it is multi-valued being a RootOf.
The user must select only the real critical points and then take the min/max of the corresponding values.

Note that the general situation is complicated even for analytic functions of a single variable because usually (including this example) the set of critical points is infinite; allvalues gives only a finite set.
RootFinding[Analytic]  finds easily many other critical points!

Edit. That's why maths is always important when using any CAS!

@Markiyan Hirnyk 

Yes, but this happens because for one of the complex critical points, the value of the function is negative, so < 9/4.

@Markiyan Hirnyk 

But {r = 1, s = 1}  is visible in your worksheet!

@Markiyan Hirnyk 

The issue here is of a different nature: extrema computed the critical points but wrongly included also the complex ones.

minimize/maximize seem to be more reliable commands for unconstrained min/max. It works for:

f:=(x*y+x*z+y*z)*(1/(x+y)^2+1/(y+z)^2+1/(z+x)^2);
g:=eval(f,[x=1,y = s^2, z = r^2]);
minimize(g,location);

@Klaroline 

Still not clear.
If you want the values of a satisfying g(c)<1 for each critical point c of g
then you get the empty set for the original function g(x)=exp(x^2*(x-a)).

You should start by explaining the exact meaning for range of "a".

@Klaroline 

@maniaerox 

That is exactly what a parametric plot is doing. Try it with your expressions.

You could add options such as:
plot( [E,T, w=0..2*Pi],labels=["E","T"], view=[-1/2 .. 2, -2..2]);

The RootOfs should come from the critical points.

f:=(a+b+c+d)^2*(a*b+a*c+a*d+b*c+b*d+c*d)^2-144*(a^2+b^2+c^2+d^2)*a*b*c*d:
g:=expand(eval(f,d=1)):
S:=[diff(g,a),diff(g,b),diff(g,c)]:
G:=Groebner[Basis](S,plex(a,b,c)):
p:=factor(G[1]);

fsolve(op(7,p));

fsolve(op(8,p));

@Preben Alsholm 
You have developped this interesting tool related to a recently posted system of ODEs.
Just curious: did you see in the past many such systems where unwanted expansions occur during dsolve?

@Axel Vogt 

He simply expressed the original symmetric function wrt other symmetric functions (those pij defined).  But he did not prove the inequality. I think that he suggested to look for other pij which could lead to success.