Proving that a critical point is a minimum I am solving a problem, which is
"Find the point of the paraboloid $P:=\{(x,y,z)\in\mathbb{R}^3 | x^2+y^2=z\}$ which is the nearest to the point $(1,1,\frac12)$."
I have already determined (using the Lagrange multipliers method) that $p=(2^{-\frac23},2^{-\frac23},2^{-\frac13})$ is a critical point of $f(x,y,z)=(x-1)^2+(y-1)^2+(z-\frac12)^2$ (which is the square of the distance function) with the restraint $g(x,y,z)=0$, where $g(x,y,z)=x^2+y^2-z$.
I need some help in order to prove that $p$ is the minimum of $f|_P$.
If possible, give me only a hint, as it is homework and I only want some idea to finish the problem by myself.

EDIT: As requested, I'm showing my calculations so far.
A critical point of $f|_P$ is, by the Lagrange multipliers method, one which satisfies $g(x,y,z)=0$ (that is, $x^2+y^2=z$) and $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ for some $\lambda \in \mathbb{R}$.
We have $\nabla f(x,y,z)=(2x-2,2y-2,2z-1)$ and $\nabla g(x,y,z)= (2x,2y,-1)$, which yields the system
$$(2x-2=\lambda 2x)\wedge(2y-2=\lambda 2y)\wedge(2z-1=-\lambda),$$
which is equivalent to
$$(x=\tfrac{1}{1-\lambda})\wedge (y=\tfrac{1}{1-\lambda}) \wedge(z=\tfrac{1-\lambda}{2}).$$
So we get $x=y$ and $2x^2=z$ (because $x^2+y^2=z$), which gives $\frac{2}{(1-\lambda)^2}=\frac{1-\lambda}{2}$, that is, $\lambda = 1-2^\frac23$.
This gives us $x=y=2^{-\frac23}$ and $z=2^{-\frac13}$.

Prof. Cook is using the multivariate index $D$ applied to the distance-squared function for points on the paraboloid measured from $(1,1,\frac{1}{2})$.  It's not the easiest thing to read, but it does establish that the point you found is at the minimal distance from the point external to the paraboloid.
Alternatively, you could put your results for $x$ and $y$ ($x = y = \frac{1}{2z}$), into the distance squared formula to reduce it to a function of $z$ alone:  
$$s^2 = (\frac{1}{2z} - 1)^2 + (\frac{1}{2z} - 1)^2 + (z - \frac{1}{2})^2  =  \frac{1}{2z^2} - \frac{2}{z} + 2 + z^2 - z + \frac{1}{4}.$$
The second derivative is $2 - \frac{4}{z^3} + \frac{3}{z^4}$, which is positive for $z = 2^{-1/3}$, so $s^2$ is minimized for that value of  $z$ (and thus for the point you found).  Since the distance $s$ is non-negative, finding the minimum for $s^2$ suffices.
EDIT: D'oh -- it would have been even easier to use $y=x$ and $z=2x^2$ to write
$$s^2 = (x - 1)^2 + (x - 1)^2 + (2x^2 - \frac{1}{2})^2  =  4x^4 - 4x + \frac{9}{4}  \Rightarrow \frac{d^2}{dx^2}(s^2) = 48x^2 , $$which is plainly positive for $x = 2^{-2/3}$; it is also easier to see that $\frac{d}{dx}(s^2) = 16x^3 - 4 = 0$ there. (I'd been working with $z$ as the principal variable the first time through...)
