Let $\alpha, \beta_1, \beta_2 \in \mathbb{R}$ such that $\alpha > 1$ and $2>\beta_1 + \beta_2$. Find the solution of the equation (if it exists)
$$2\beta_1(\alpha-1)(2-(\beta_1+\beta_2))\geq \alpha^2$$
I don't think so the solution exists, if it does any help will be appreciated. And yes $\beta_1, \beta_2 >0$
On
Just another way of showing there are no solutions:
If possible, let $2\beta_1(\alpha-1)(2-\beta_1-\beta_2)\geqslant\alpha^2$.
$\implies 2\beta_1(\alpha-1)(2-\beta_1)>\alpha^2$
By AM-GM, $\beta_1(2-\beta_1)\leqslant1$, so we must have:
$2(\alpha-1)>\alpha^2 \implies -1>(\alpha-1)^2$
which is clearly false.
There are no triples of real numbers $(\alpha,\beta_1,\beta_2)$ such that $$\alpha\gt 1\quad\text{and}\quad 2\gt\beta_1+\beta_2\quad\text{and}\quad \beta_1\gt 0\quad\text{and}\quad \beta_2\gt 0$$ $$2\beta_1(\alpha-1)(2-(\beta_1+\beta_2))\geq \alpha^2\tag1$$ $(1)$ is equivalent to $$\alpha^2-2\beta_1(2-\beta_1-\beta_2)\alpha+2\beta_1(2-\beta_1-\beta_2)\le 0\tag2$$
Seeing $(2)$ as a quadratic inequality on $\alpha$, we have, using AM-GM inequality, $$\begin{align}&4\beta_1^2(2-\beta_1-\beta_2)^2-4\cdot 2\beta_1(2-\beta_1-\beta_2)\\\\&=-4\beta_1^2(2-\beta_1-\beta_2)\left(\beta_2-2+\beta_1+\frac{2}{\beta_1}\right)\\\\&\le -4\beta_1^2(2-\beta_1-\beta_2)\left(\beta_2-2+2\sqrt{\beta_1\cdot\frac{2}{\beta_1}}\right)\\\\&=-4\beta_1^2(2-\beta_1-\beta_2)\left(\beta_2-2+2\sqrt{2}\right)\\\\&\lt 0\end{align}$$ we see that $(2)$ has no real solutions.