Suppose $WLOG f\geq g\geq h$. Then, $g+h >f$ or $g+h - 1\geq f$ But According the the second condition, $f-1 \geq g-1 + h-1 $. This means $f \geq g + h - 1 $. Thus equality holds and $f=g+h-1$ Or, $h+g-f =1$.Which is a constant

How do you obtain $g+h-1 \geq f$? It is not like they can be only integers, it is true for all reals.

Are we forgetting that the WLOG didn't make much sense either? You said $h+g-f$ is constant for some fixed value of $x$ as the WLOG can be done for only a fixed value of $x$. If done for the whole domain, generality is actually lost.

Let us take a very large value of $x$, say $x>N$ where $N>>>$ the coefficients of the polynomial. In that case, as the polynomials are quadratic, one can assume that $f(x)>g(x)$ is the same as the coefficients of the square term of $f$ being larger than $g$ for $x>N$. In that sense, assume WLOG that $f(x) \geq g(x) \geq h(x)$ for all sufficiently large $x$. Hence, from the conditions, we can see that $g+h>f$ and $g+h-1\leq f$ for all $x>n$. That is, the polynomial $g+h-f$ is bounded between $0$ and $1$ for all sufficiently large reals $x$, and one can easily prove that $g+h-f$ has to be a constant between $0$ and $1$ for this to happen.

Very tiny correction: All trinomials aren't quadratics. But still the assumption that the highest power of $x$ is the same does hold trivially. @below sorry, my bad

It said "quadratic trinomials" in the problem.