Consider the following system of polynomial equations in $x_1,\ldots,x_6$:
\begin{eqnarray*} x_1x_2 - α & = & 0\\ x_3x_4 - β & = & 0\\ x_5x_6 - γ & = & 0\\ x_1x_4+x_3x_2 - δ & = & 0\\ x_1x_6+x_5x_2 - ε & = & 0\\ x_3x_6+x_5x_4 - ζ & = & 0 \end{eqnarray*} where all constants are strictly positive.
Now assume that $x_1= c>0$, which makes the system overdetermined.
Suppose also that real solutions to the resulting determined system exist when any of the last three equations is removed from it.
Where should I look for conditions that ensure that real solutions exist for the original system?
EDIT: I may have made a typo my Maple in my answer yesterday, because I get a different answer today.
In a "plex" Groebner basis for your system found by Maple (with variable order $x_6, x_5, x_4, x_3, x_2, x_1, \alpha,\beta,\gamma,\delta,\epsilon,\zeta$), the first polynomial (which is the only one that doesn't contain any $x$'s) is $$ 4\,\alpha\,\beta\,\gamma-\alpha\,{\zeta}^{2}-\beta\,{\epsilon}^{2}-{ \delta}^{2}\gamma+\delta\,\epsilon\,\zeta $$ So that being $0$ is a necessary condition for a solution to exist. It is not sufficient though: the third polynomial in the Groebner basis is $$ \left( 4\,{\beta}^{2}\gamma-\beta\,{\zeta}^{2} \right) {x_{{1}}}^{2}+ \left( -4\,\beta\,\delta\,\gamma+\delta\,{\zeta}^{2} \right) x_{{1}}x _{{3}}+ \left( \beta\,{\epsilon}^{2}+{\delta}^{2}\gamma-\delta\, \epsilon\,\zeta \right) {x_{{3}}}^{2} $$ and unless the discriminant of this (as a quadratic in $x_3$) is nonnegative there will be no real solutions. For example, with $\alpha=\beta=\gamma=\delta=\zeta=1$ and $\epsilon=2$, the first polynomial is $0$ but there are no real solutions, as the third polynomial is $3 x_1^2 - 3 x_1 x_3 + 3 x_3^2$ with discriminant $-27$.
Note, by the way, that each term involving $x$'s in your system is the product of an odd-indexed and an even-indexed $x$. Thus the system is invariant under the transformations $$x_1 \to t x_1,\ x_3 \to t x_3,\ x_5 \to t x_5,\ x_2 \to x_2/t,\ x_4 \to x_4/t,\ x_6 \to x_6/t$$ for $t \ne 0$. In particular, specifying a nonzero value for $x_1$ will not make a difference.