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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?

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    Since $f$ is nonconvex you may not use duality to minimise it (you do not have strong duality and the dual problem is not equivalent to the primal one). Furthermore, notice that $f$ is not defined for $x=0$ - unless you assign it the value $f(0) = +\infty$. If you need to prove that $(1,0)$ is indeed a (local) minimiser, you need to take the first and second-order optimality conditions. – 2017-01-10
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    Thanks. What do you mean by first and second order optimal conditions. $(1,0)$ should be also a global minimizer. How can I prove that. – 2017-01-10
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    @PantelisSopasakis Also, it seems that the Hessian is semi positive definite, then why it is not convex? – 2017-01-10
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    I'm sorry, my mistake - indeed, $f$ is convex. Then, your procedure using duality is correct. You may make use of the KKT conditions. – 2017-01-11

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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 $0\leq x\leq1$.

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    If for example $\alpha=0.9$ then $x=18$, which is not supported. – 2017-01-10
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    The condition $x\geq0$, $y\geq0$ is not anywhere in the formulas. You cannot expect the formulas to know about it. If the solution is out of range, it means that there is not critical point within that range. – 2017-01-10
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    So then it means that the solution must be on the boundary, right? – 2017-01-11
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    Yes. $\ \ \ \ \ \ $ – 2017-01-11