I'm given the function
$$ f(x,y) = \alpha\cdot\frac{2-x}{x}+(1-\alpha)\cdot(2-y) $$ which I want to minimize with respect to $0\leq x,0\leq y$, under the constraint $x+y=1$. It is also known that $1/2\leq\alpha$. I know that the optimizer is given by $(x,y)=(1,0)$.
We define the Lagrangian: $$ L(x,y,\lambda) = \alpha\cdot\frac{2-x}{x}+(1-\alpha)\cdot(2-y)+\lambda\cdot(x+y-1) $$
Then, taking derivative with respect to $x$ and $y$ we get $$ \alpha\frac{2}{x^2}=\lambda $$ and $$ (1-\alpha)=\lambda $$ which does not make sense. Where is my mistake?
No mistake. Your equations give, provided that $\alpha<1$, $$ x=\left(\frac{2\alpha}{1-\alpha}\right)^{1/2},\ \ \ \ \ \ \ y=1-x=1-\left(\frac{2\alpha}{1-\alpha}\right)^{1/2} $$ One would then need adequate choices of $\alpha$ to have $0\leq x\leq1$.