In my textbook I have a following problem:
Find extremas of the function $$f(x, y, z) = \frac{\alpha}{x} + \frac{\beta}{y} + \frac{\gamma}{z}$$ subject to $ax + by + cz = 1$, where $a, b, c, \alpha, \beta, \gamma > 0$ and $x, y, z$ have the same sign.
The Lagrange function is of the form $$\Lambda = \frac{\alpha}{x} + \frac{\beta}{y} + \frac{\gamma}{z} + \lambda\left(ax + by + cz - 1 \right).$$ Differentiating it w.r.t. $x$, $y$ and $z$ and putting it equal to the zero we get: $$\partial_x \Lambda = -\frac{\alpha}{x^2} + \lambda a = 0,$$ $$\partial_y \Lambda = -\frac{\beta}{y^2} + \lambda b = 0,$$ $$\partial_z \Lambda = -\frac{\gamma}{z^2} + \lambda c = 0.$$ Using that we can find that $a = \frac{\alpha}{\lambda x^2}$, $b = \frac{\beta}{\lambda x^2}$ and $c = \frac{\gamma}{\lambda z^2}$. Putting this to the equation for constraint we find that $$\lambda = \frac{\alpha}{x} + \frac{\beta}{y} + \frac{\gamma}{z}.$$
From now on I'm lost. I've tried to substitute Lagrange multiplier into derivatives but I get systems of the equations which I have no idea how to solve. What might be the trick?
Assuming $x > 0, y > 0, z > 0$ we have
$$ a x + b y + z c \ge 3 \sqrt[3]{a b c x y z} $$
note that there exist $\bar x, \bar y \bar z$ such that
$$ \sqrt[3]{a b c \bar x \bar y \bar z} = 1 $$
and for these $\bar x, \bar y \bar z$ we have also
$$ \frac{\alpha}{\bar x}+\frac{\beta}{\bar y} +\frac{\gamma}{\bar z}\ge 3\sqrt[3]{\frac{\alpha\beta\gamma}{\bar x\bar y\bar z}}3\sqrt[3]{abc \bar x\bar y\bar z} = 9\sqrt[3]{\alpha\beta\gamma abc} $$
so the sought minimum is $ 9\sqrt[3]{\alpha\beta\gamma abc}$
Analogous procedure if $x < 0, y < 0, z < 0$