What does w = (u,v) mean?

I've read your worksheet, and I understand the problem. But could you make a smaller example of it not working? I am measuring by the number of substitutions. So, could you make an example with, say, three substitutions, where the iterative application of simplify with side relations does not yield the same results as the all-at-once application?

Note to other readers: I am not saying that the iterative application should give the same results as the all-at-once! However, in an earlier very similar problem that Casper had, I was able to order the side relations in such a way that the results were the same, with the iterative way being thousands of times faster.

What exactly do you mean by Components? Do you mean embedded components? I can offer a great deal of advice about your worksheet if that's what's wanted.

(I am tempted to use my editorial discretion to make this a Question, but this time I'll leave it to someone else to make the final decision.)

Suppose we take the circle as fixed, b = c, and a < < b. As a --> 0, b and c approach a diameter of the circumcircle. So, b --> 2*R and c --> 2*R. Thus R*p = R*(a+b+c)/2 --> 2*R^2 and 2*a*R + b*c --> 4*R^2. So, R*p/(2*a*R+b*c) --> 1/2.

OTOH, suppose the circle is fixed, b = c, and a is at its maximal value, 2*R. Then it's a right triangle and b = c = sqrt(2)*R. Then R*p/(2*a*R+b*c) = (1+sqrt(2))/6 = 0.402, which is just slightly larger than 2/5.

While neither of these arguments is a proof of either side of the inequality, they do make it seem plausible. Somewhere we need to use the acuteness condition. Is acuteness equivalent to  a^2 < b^2 + c^2 with a being the longest side? Is acuteness equivalent to the center of the circle being inside the triangle?

How about trying to use Maple in the c# program? If you get an error, then trap it.

@Alejandro Jakubi FWIW, lprint shows that the 2D if in prefix form is translated to 1D `if` as an automatic simplification.
 

@Alejandro Jakubi FWIW, lprint shows that the 2D if in prefix form is translated to 1D `if` as an automatic simplification.
 

@Andriy It is allowed without the quotes in 2d input. In 1d input (Maple Input), it is an immediate syntax error to use if in prefix form without the quotes, because if is a ?keyword . The prefix form `if` means the same thing as if it were defined thus

`if`:= proc(condition, expression1::uneval, expression2::uneval)
local B;
     B:= evalb(condition);
     if B=true or B=false then
          if B then  return eval(expression1)  else  return eval(expression2)  end if
     else
          return '`if`(condition, expression1, expression2)'  #return unevaluated
     fi
end proc

But, it is a builtin, or kernel, procedure, so it isn't actually written in Maple code as above.    

@Andriy It is allowed without the quotes in 2d input. In 1d input (Maple Input), it is an immediate syntax error to use if in prefix form without the quotes, because if is a ?keyword . The prefix form `if` means the same thing as if it were defined thus

`if`:= proc(condition, expression1::uneval, expression2::uneval)
local B;
     B:= evalb(condition);
     if B=true or B=false then
          if B then  return eval(expression1)  else  return eval(expression2)  end if
     else
          return '`if`(condition, expression1, expression2)'  #return unevaluated
     fi
end proc

But, it is a builtin, or kernel, procedure, so it isn't actually written in Maple code as above.    

@Alejandro Jakubi Very nice trick: computing series with respect to two different variables in the same computation.

@Alejandro Jakubi Very nice trick: computing series with respect to two different variables in the same computation.

Perhaps the information that you seek can be found on the help pages ?pdsolve,numeric,pdemethods and ?pdsolve,numeric,education .

@yangtheary There is no aspect of the problem statement or diagram that distinguishes the up-down dimension from the left-right dimension. Therefore, the outer lattice and the inner lattice are each squares. Therefore, the answer is a sum of two squares.

@yangtheary There is no aspect of the problem statement or diagram that distinguishes the up-down dimension from the left-right dimension. Therefore, the outer lattice and the inner lattice are each squares. Therefore, the answer is a sum of two squares.

@casperyc I haven't looked at it yet, but when you say "failed" do you mean that the answer is wrong or that it takes too long to get?

The reverse order trick only worked becuase of a special form of the relations: Each one was the previous one essentially to a higher power (except that first one was independent of the rest). In general, if the relations are applied sequentially, the results will be wrong, i.e., not as simplified as if they were applied all together.

Please make this a separate Question thread. It is some trouble to dig up old threads.