Optimization with a constrained function Okay so I understand how to find points of extrema when for example, 
We have $3x^2 + 2y^2 + 6z^2$ subject to the constaint $x+y+z=1$. I followed the method of the Lagrange multiplier and resulted in the point $(\frac13, \frac12, \frac16)$. I know this point is correct. 
However, I don't know how to prove that it's a minimum value. Any ideas?

An elementary proof would be as follows: notice that by "completing the square",
$$3x^2 - 2x + 2y^2 - 2y + 6z^2 - 2z \\
= 3\left(\left(x - \frac{2}{6}\right)^2 - \left(\frac{2}{6}\right)^2\right) + 2\left(\left(y - \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2\right) + 6\left(\left(z - \frac{2}{12}\right)^2 - \left(\frac{2}{12}\right)^2\right)\\
= 3\left(x - \frac{2}{6}\right)^2 + 2\left(y - \frac{1}{2}\right)^2 + 6\left(z - \frac{2}{12}\right)^2 - \frac{1}{3} - \frac{1}{2} - \frac{1}{6}\\
\ge -1$$
(In case you're confused: $-\frac{1}{3} - \frac{1}{2} - \frac{1}{6} = -1$)
But $x + y + z = 1$, so we have
$$3x^2 + 2y^2 + 6z^2 -2 \ge -1$$
$$3x^2 + 2y^2 + 6z^2  \ge 1$$
It is quite evident (from the first inequality) that equality is achieved when $x = \frac{1}{3}, y = \frac{1}{2}, z = \frac{1}{6}$.

A more advanced method: using the Cauchy-Schwarz inequality, we have
$$(x + y + z)^2 \le \left(3x^2 + 2y^2 + 6z^2\right)\left(\frac{1}{3} + \frac{1}{2} + \frac{1}{6}\right)$$
But $x + y + z = 1$, so we can substitute this and simplify a bit to get:
$$3x^2 + 2y^2 + 6z^2 \ge 1$$
Equality holds when $3x = 2y = 6z$. From this, it can be trivially deduced that indeed, $(x,y,z) = \left(\frac{1}{3}, \frac{1}{2}, \frac{1}{6}\right)$ minimizes the expression  with the given constrain.
While not exactly nor directly relevant to Lagrange Multipliers, this does provides a simpler alternative for certain optimization problems.
