Are there three real rational functions $f,g,h$ with no poles in $[0,1]$, such that $f\geq 0,g\geq 0,f+g \geq h \geq 0$ on $[0,1]$ and the curve $\gamma(t)=(f(t),g(t),h(t)) (t\in [0,1])$ passes through each of the four points $(1,0,0),(0,1,0),(1,0,1),(0,1,1)$ ?
My guess is that it is not possible because those points are "sharp boundary points" of the set in which the curve $\gamma$ must stay.
Yes, there are. Consider $$\begin{align} \gamma(0) &= (1, 0, 0)\\ \gamma(1/4) &= (0, 1, 1)\\ \gamma(3/4) &= (1, 0, 1)\\ \gamma(1) &= (0, 1, 0)\end{align}$$ where $$\begin{align} f(x) &= -16 x^3 + 24 x^2 - 9 x + 1\\ g(x) &= 16 x^3 - 24 x^2 + 9 x\\ h(x) &= \frac{8192}{27} x^6 - \frac{8192}{9} x^5 + \frac{8960}{9} x^4 - \frac{12800}{27} x^3 + \frac{256}{3} x^2\end{align}$$
For $0 \le x \le 1$, $$\begin{align} 1 \ge f(x) &\ge 0 \\ 1 \ge g(x) &\ge 0 \\ 1 \ge h(x) &\ge 0 \\ f(x) + g(x) &= 1 \end{align}$$