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In a recent post, the following inequality was proved with Maple:



(a,b,c,d >= 0).

Here is another direct proof attempt.

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:=minimize(g, [a=0..infinity,b=0..infinity,c=0..infinity]):
length(s);   # huge
        304856
map(evalf@evalf[500],s);

map(evalf@evalf[1000],s);


So, Maple returns the expression

min(0,r1,r2,r3)


where r1,r2,r3 are huge expressions containing RootOfs. In order to evaluate them, several hundreds of digits are needed.
The solution seems to be correct, but the question is: may we (mathematically) accept it? What do you think?

 

eulermac(1/(n*ln(n)^2),n=2..N,1);  #Error
Error, (in SumTools:-DefiniteSum:-ClosedForm) summand is singular in the interval of summation


eulermac(1/(n*ln(n)^2+1),n=2..N,1);  #nonsense

 

 

The well known William Lowell Putnam Mathematical Competition (76th edition)  took place this month.
Here is a Maple approach for two of the problems.

1. For each real number x, 0 <= x < 1, let f(x) be the sum of  1/2^n  where n runs through all positive integers for which floor(n*x) is even.
Find the infimum of  f.
(Putnam 2015, A4 problem)

f:=proc(x,N:=100)
local n, s:=0;
for n to N do
  if type(floor(n*x),even) then s:=s+2^(-n) fi;
  #if floor(n*x) mod 2 = 0  then s:=s+2^(-n) fi;
od;
evalf(s);
#s
end;

plot(f, 0..0.9999);

 

min([seq(f(t), t=0.. 0.998,0.0001)]);

        0.5714285714

identify(%);

So, the infimum is 4/7.
Of course, this is not a rigorous solution, even if the result is correct. But it is a valuable hint.
I am not sure if in the near future, a CAS will be able to provide acceptable solutions for such problems.

2. If the function f  is three times differentiable and  has at least five distinct real zeros,
then f + 6f' + 12f'' + 8f''' has at least two distinct real zeros.
(Putnam 2015, B1 problem)

restart;
F := f + 6*D(f) + 12*(D@@2)(f) + 8*(D@@3)(f);

dsolve(F(x)=u(x),f(x));

We are sugested to consider

g:=f(x)*exp(x/2):
g3:=diff(g, x$3);

simplify(g3*8*exp(-x/2));

So, F(x) = k(x) * g3 = k(x) * g'''
g  has 5 distinct zeros implies g''' and hence F have 5-3=2 distinct zeros, q.e.d.

 

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