EDIT#2: Here:
http://www.animations.physics.unsw.edu.au/jw/images/AC_files/resonancev(t).gif
is an image of the waveforms in this circuit.
You can't simply add the magnitudes of voltage drops across each element to get the maximum voltage. I am not entirely sure, but it looks like in your first equations you are simply writing the magnitudes of VC, VL and VR. You have to incorporate the phase of each. That's why your second method looks correct to me.
Think of it this way. If you are forming a right triangle that has sides, a, b, c such that:
$$ c^2 = a^2 + b^2 $$
If we look at sides a and b as vectors from the origin then we can't simply say that the sum of the magnitudes of a and b is c.
$$ c \ne a + b $$
We can say, however, that:
$$ \vec{c} = \vec{a} + \vec{b} $$
Use this analogy to think about your circuit. Inductors and capacitor have voltage phasors with angles:
$$ \vec{V_L} = IZ = j(I\omega L) = I\omega L\angle 90^\circ $$
$$ \vec{V_C} = IZ = \frac{I}{j\omega C} = -j\frac{I}{\omega C} = \frac{I}{\omega C}\angle{-}90^\circ $$
$$ \vec{V_R} = IZ = IR $$
Then we can get the vector sum, and take its magnitude:
$$ \vec{V_{total}} = \vec{V_L} + \vec{V_C} + \vec{V_R} $$
$$ \vec{\left\lvert V_{total} \right\lvert} = \left\lvert \vec{V_L} + \vec{V_C} + \vec{V_R} \right\lvert $$
In short, the angles are important and because part of the phasor sum.
EDIT: added vector notation for clarity with phasors