Minimize $f(x, y, z) = \frac{a}{x} + \frac{b}{y} +\frac{c}{z}$ with $x+y+z=1$ Problem:
Minimize the function $$f(x, y, z) = \frac{a}{x} + \frac{b}{y} +\frac{c}{z}$$ where $a, b, c$ are constants and $a, b, c, x, y, z > 0$.
In addition, $$x+y+z=1$$.
I have a solution, but I'm not sure if it's right - in particular, what can I do to check it? (I can't plot something like this, and I can't seem to make Wolfram accept $a, b, c$ as constants)
Solution:
$$f(x, y, z) = f(x, y) = \frac{a}{x} + \frac{b}{y} +\frac{c}{1 - x - y}$$
Now finding when the partial derivatives vanish will give the turning points.
$$f_x = -\frac{a}{x^2} + \frac{c}{(1-x-y)^2}$$
$$f_y = -\frac{b}{y^2} + \frac{c}{(1-x-y)^2}$$
If $f_x = f_y = 0$, then
$$\frac{a}{x^2} = \frac{b}{y^2} = \frac{c}{(1-x-y)^2} = \frac{c}{z^2}$$
So
$$\begin{cases}
x = \frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\\
y = \frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\\
z = \frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}
\end{cases}$$
and
$f_{min}=(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$
 A: Another way to do the problem is to use Lagrangian optimisation. Then you would try to minimize the Lagrangian $$L(x,y,z,\lambda)=f(x,y,z)-\lambda(x+y+z-1)$$
Computing partial derivatives gives: $$\frac{\partial L}{\partial x}=-\frac a{x^2}-\lambda\\\frac{\partial L}{\partial y}=-\frac b{y^2}-\lambda\\\frac{\partial L}{\partial z}=-\frac{c}{z^2}-\lambda\\\frac{\partial L}{\partial \lambda}=-(x+y+z-1)$$
We require these to all be $0$. The fourth equation is then just the constraint, while the others say the minimum occurs at $$(x,y,z)=\left(\sqrt{-\frac a\lambda},\sqrt{-\frac b\lambda},\sqrt{-\frac c\lambda}\right)$$
Substituting this into the constraint gives $$\sqrt{-\frac a\lambda}+\sqrt{-\frac b\lambda}+\sqrt{-\frac c\lambda}=1$$ We can solve this to give $$\lambda=-\left(\sqrt a+\sqrt b+\sqrt c\right)^2$$And so the solution to the problem is the same as what you got.
You can then use theh Bordered Hessian method to determine if it's a minimum or maximum.

Note, if you wanted to check this some other way, you wouldn't have to show $\forall x,y,z,a,b,c>0$ $$\frac ax+\frac by+\frac cz\ge\left(\sqrt a+\sqrt b+\sqrt c\right)^2$$ since this loses sight of the constraint $x+y+z=1$. Without the constraint, we can just take $x,y,z$ to be very large, and then the LHS is clearly smaller than the RHS. 
In fact, what needs to be verified is the above inequality $\forall a,b,c>0$ and $\forall (x,y,z)\in P$ where $P$ is the part of the 2-D plane $x+y+z=1$ that is in the region $x,y,z>0$. 
It is hard to capture this constraint correctly by simply substituting something in. The best way to try to prove this is to find the minimum of the LHS minus the RHS. This is the same optimisation problem as before, and is most easily solved / proven in the same way as before, with a Lagrangian. 
I say this to reassure you that there's no need to be "more sure" about the answer being correct - if it is what you get from minimising the function while considering the constraint, then it is correct. Proving such inequalities is a lot more difficult and not necessary.
