If 3 numbers of the form $\frac{1}{q^2+1}$ have a mean of $\frac{1}{2}$, will one of them always be $\frac{1}{2}$?

Question

I recently came onto a set of numbers with an interesting pattern, but I can't prove it always holds true. I'd appreciate any help in figuring it out.

Let the set $S$ $$S = \left \{ \frac{1}{q^2+1} , q \in \mathbb{Q^+} \right \}$$

And three elements of that set $a,b,c \in S$

My conjecture is that if their average is equal to $\frac{1}{2}$, then at least one of them must be equal to $\frac{1}{2}$.

$$\frac{a+b+c}{3} = \frac{1}{2} \implies (a =\frac{1}{2}) \lor (b=\frac{1}{2}) \lor (c=\frac{1}{2})$$

Do you have any idea on how to prove, or disprove that claim?

Edit to provide examples: You can easily generate sets of numbers that verify the conjectured pattern. Let $f(q) = \frac{1}{q^2+1}$ for rational $q$.

All triplets of the form $(f(1), f(q), f(\frac{1}{q}))$ have an average of $\frac{1}{2}$.

Answer 1

Unfortunately, the postulated conjecture is not true:

Indeed, examples of admissible $q$-triples that disprove the statement are

$$ (\tfrac{3}{2}, \tfrac{2}{9},\tfrac{41}{23})\\ (\tfrac{9}{2}, \tfrac{2}{3},\tfrac{23}{41}) $$

I found those numbers by just brute force looking for them. I did this by iterating through rational numbers $a,b$, then I calculate $c$ such that $f(a)+f(b)+f(c)=\tfrac{3}{2}$ and then check if $c$ is rational. Please have a look at the following Maple code

restart:

f:=x->1/(1+x^2);

N:=10;

for p1 from 2 to N do
  for q1 from 1 to p1-1 do
    for p2 from 1 to N do
      for q2 from p2+1 to N do
        
        a:=p1/q1;
        b:=p2/q2;
        R:=3/2-f(a)-f(b);
        c:=sqrt(1/R-1);

        if( type(c, rational) and R<>1/2 ) then 
          print(a,b,c,f(a),f(b),f(c),f(a)+f(b)+f(c)) 
        end if;

      end do;
    end do;
  end do;
end do;

Increasing $N$ in the code produces more and more triplets that violate the conjecture.

Answer 2

$\frac{1}{1+x^2}+\frac{1}{1+y^2}=\frac{x^2+y^2+2}{(1+x^2)(1+y^2)}$

$\frac{3}{2}-\frac{x^2+y^2+2}{(1+x^2)(1+y^2)}=\frac{3(1+x^2)(1+y^2)-2(x^2+y^2+2)}{2(1+x^2)(1+y^2)}$

$=\frac{3+3x^2+3y^2+3x^2y^2-2x^2-2y^2-4}{2(1+x^2)(1+y^2)}=\frac{x^2+y^2+3x^2y^2-1}{2(1+x^2)(1+y^2)}$

$\frac{1}{1+z^2}=\frac{x^2+y^2+3x^2y^2-1}{2(1+x^2)(1+y^2)}$

$2(1+x^2)(1+y^2)=(1+z^2)[(1+x^2)(1+y^2)+2x^2y^2-2]$

The ratios $x,y,z$ satisfy this relationship: $(1-z^2)(1+x^2)(1+y^2)=2(1+z^2)(x^2y^2-1)$.

Further, this relationship holds when pair wise swapping any of the ratios.

So as pointed out, if any 1 of them is 1, the other two must be reciprocals.

$(1+x^2)(1+y^2)-2(x^2y^2-1)=2z^2(x^2y^2-1)+z^2(1+x^2)(1+y^2)$

$z^2=\frac{(1+x^2)(1+y^2)-2(x^2y^2-1)}{2(x^2y^2-1)+(1+x^2)(1+y^2)}$

So long as $xy\ne1$, $z\ne 1\to \frac{1}{1+z^2}\ne 1/2$. One still needs to demosntrate existence.

Let $x=c/d, y=e/f$ and $gcd(c,d)=gcd(e,f)=1$

$\frac{p^2}{q^2}=\frac{(c^2+d^2)(e^2+f^2)-2(c^2e^2-d^2f^2)}{2(c^2e^2-d^2f^2)+(c^2+d^2)(e^2+f^2)}$

$2p^2(c^2e^2-d^2f^2)+p^2(c^2+d^2)(e^2+f^2)=q^2(c^2+d^2)(e^2+f^2)-2q^2(c^2e^2-d^2f^2)$

$2(p^2+q^2)(c^2e^2-d^2f^2)+(p^2-q^2)(c^2+d^2)(e^2+f^2)=0$

$2(p^2+q^2)(d^2f^2-c^2e^2)=(p^2-q^2)(c^2+d^2)(e^2+f^2)$