If $\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$ for real $t$ and $0
$$\frac{t^2-x^2}{(t^2+x^2)^2}+\frac{(1-x)^2-t^2}{((1-x)^2+t^2)^2}=0$$ 
  where $0<x<1$ and $t\in\mathbb{R}$. Prove that $x=1/2$. 
It is evident that $x=1/2$ satisfies the above equation. Please help.

Seems the fastest way to show this is indeed considering the polynomial left after clearing the denominator and dividing by $2(x-\frac{1}{2})$
$$f(x)=-3 t^4 - 2 t^2 x^2 + 2 t^2 x - t^2 + x^4 - 2 x^3 + x^2$$
Considering $$f'(x)= -2 (2 x - 1) (t^2 - x^2 + x)$$ we see that as $(t^2+x(1-x))>0$ for $0<x<1$ thus the function increases on $\left(0,\frac{1}{2}\right)$ and decreases on $\left(\frac{1}{2},1\right)$ thus we have to check only that $f(0),\,f\left(\frac12\right),\,f(1)$ has the same sign, indeed:
$$f(0)=-3t^4-t^2\le 0$$ $$f(1)=-3t^4-t^2\le 0$$
$$f\left(\frac{1}{2}\right)=-3 t^4 - \frac{t^2}{2} + \frac{1}{16}$$
But the latter expression factors as
$$-\frac{1}{16} (4 t^2 + 1) (12 t^2 - 1)$$
so for $12t^2-1<0$ the statement is not true.
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