I was trying to prove that the following systems of non-linear equations have no solution in a given domain in the past a few weeks. But has no clue. The problem is the following: prove \begin{equation*} \begin{cases} \frac{2(1-\alpha)^3}{3(1-\beta)^3}+\frac{(2-\alpha)^3}{3(2-\beta)^2}+\frac{\alpha(1-\alpha)^2}{1-\beta}-\frac{\alpha(2-\alpha)^2}{2(2-\beta)}-\frac{(1-\alpha)^2(2+\alpha)}{3(1-\beta)^2}=0,\\ \frac{(2-\alpha)^3}{12(2-\beta)^3}\left(5\alpha-2-\beta(3\alpha+2)/2\right)-\frac{(1-\alpha)^3}{12(1-\beta)^4}\left(\beta^2(3\alpha+1)+4\beta(1-\alpha)-5\alpha+1\right)=0. \end{cases} \end{equation*}
has no solutions in $0\leq\beta\leq\alpha\leq 1$.
I can numerically verify that this claim is true. I can even use matlab symbolic calculation tool box and it returns that there is indeed no solution in $0\leq\beta\leq\alpha\leq 1$. (But matlab symbolic calculation does not output the steps).
The algebra of these expressions are not very friendly. High orders of $\alpha$ and $\beta$. Taking derivatives does not help. And both the planes (the 1st and the 2nd expressions) have positive and negative part.
I really don't know if this claim can be proved rigorously. Any comments and/or suggestions are highly appreciated!
I have no experience at all with Matlab but I can suppose that, $f(\alpha,\beta)$ and $g(\alpha,\beta)$ being the rhs, it tries to minimize (under the given inequality constraints) the function $$\Phi(\alpha,\beta)=f^2(\alpha,\beta)+g^2(\alpha,\beta)$$ hoping to find $\Phi_{min}=0$ (this would mean that, at least, one solution exists.
For your problem the result of such optimization is $$\alpha=0.494090 \qquad \beta=0 \qquad \Phi_{min}=0.00108287$$
We can check for $\beta=0$.
$$f(\alpha,0)=-\frac{1}{6} (a-2)^2 (2 a-1)$$ $$g(\alpha,0)=\frac{1}{96} \left(-45 a^4+160 a^3-216 a^2+128 a-24\right)$$ but $$9216\,\Phi(\alpha,\beta)=2025 a^8-14400 a^7+46064 a^6-89856 a^5+$$ $$122800 a^4-122368 a^3+82048 a^2-30720 a+4672$$ has no real root but its minimum is what was given above.