A quicker way to do a Lagrange multiplier problem I was working on the problem: minimize $x + 4z$ subject to $x^2 + y^2 + z^2 \le 2 $.  I have it solved, I want a faster method for use in standardized exams.
My work:
I tackled this using Lagrange Multipliers, considering the interior by looking for points where all individual partial derivatives of $x+ 4z$ are zero, (of which there are none).  Then considering the boundary 
$$ (x + 4z) - \lambda ( x^2 + y^2 +z^2 - 2) $$ 
From here I differentiated w.r.t x,y,z, $\lambda$ and set equal to 0 to yield 
$$ 1 - 2\lambda x = 0 \rightarrow 1 = 2\lambda x $$ 
$$ - 2 \lambda y = 0  \rightarrow 0 = 2 \lambda y \rightarrow y=0$$
$$ 4 - 2 \lambda z = 0 \rightarrow 4 = 2\lambda z$$
$$  - (x^2 + y^2 +z^2  -2 ) = 0  \rightarrow x^2 +z^2 = 2$$
Looking at equations 1, 3 we have 
$$ \frac{1}{2} = \lambda x, 2 = \lambda z $$ And therefore 
$$ \frac{1}{4} + 4 = \lambda^2 (x^2 +z^2 ) = 2 \lambda ^2 $$ 
$$ \frac{17}{8} = \lambda ^2 $$ 
And thus $$ \lambda = \pm \sqrt{ \frac{17}{8} } $$ 
$x = \frac{1}{2\lambda}, z = \frac{2}{\lambda} $
Yields 
$$ x + 4z = \frac{1}{2\lambda} + 4 \frac{2}{\lambda} = \frac{17}{2 \lambda} 
 = \pm \sqrt{17} \sqrt{2} = \pm \sqrt{34}$$ 
Clearly $-\sqrt{34}$ is smaller, so we opt for that as our solution.
Now while this works, and makes sense, its not satisfactory as it TAKES SO LONG. And on a Math GRE where the expectation is to do this under 30 seconds a problem, I was hoping there was a faster method. Any suggestions? [Also open to ways to speed up the process, since even the same method with a different angle might be superior] 

By Cauchy-Schwarz,
$$|x + 4z| \le \sqrt{1^2 + 4^2} \sqrt{x^2+z^2} \le \sqrt{34}.$$
Then think about when Cauchy-Schwarz attains equality.
