So I think I have all the pieces, but I lack your brain power.
Minkowski's Inequality is proved (in Rudin's text) via Holder's Inequality, namely
For [itex]\frac{1}{p}+\frac{1}{q}=1[/itex], [itex]1< p< \infty,1<q< \infty,[/itex], and let f and g be measurable functions on X with range in [itex]\left[ 0, \infty\right][/itex]. Then
[tex]\left\{\int_X fg d\mu\right\}\leq \left\{\int_X f^{p} d\mu\right\} ^{\frac{1}{p}}\left\{\int_X g^{q} d\mu\right\} ^{\frac{1}{q}}[/tex]
To prove Minkowski's put [itex](f+g)^{p}=f(f+g)^{p-1}+g(f+g)^{p-1}[/itex] and apply Holder's inequality to each of the terms on the right, that is
[tex]\left\{\int_X f(f+g)^{p-1} d\mu\right\}\leq \left\{\int_X f^{p} d\mu\right\} ^{\frac{1}{p}}\left\{\int_X (f+g)^{(p-1)q} d\mu\right\} ^{\frac{1}{q}}[/tex]
and
[tex]\left\{\int_X g(f+g)^{p-1} d\mu\right\}\leq \left\{\int_X g^{p} d\mu\right\} ^{\frac{1}{p}}\left\{\int_X (f+g)^{(p-1)q} d\mu\right\} ^{\frac{1}{q}}[/tex]
note that (p-1)q=p and that 1/q=1-1/p then add the above inequalities to get Minkowski's inequality in the following form:
[tex]\left\{\int_X \left( f+g\right) ^{p} d\mu\right\} ^{\frac{1}{p}} \leq \left\{\int_X f^{p} d\mu\right\} ^{\frac{1}{p}} + \left\{\int_X g^{p} d\mu\right\} ^{\frac{1}{p}}[/tex]
for non-negative f and g. Equality holds in Holder's inequality if, and only if, there are constants a and b such that [itex]af^p=bg^q[/itex].
To arrive at the form of Minkowski's inequality I have posted above, note that
[tex]\left| f + g \right| \leq \left| f \right| + \left| g \right| \Rightarrow \left| f + g \right| ^{p} \leq \left( \left| f \right| + \left| g \right| \right) ^{p}[/tex]
by the triangle inequality. For complex f and g, their magnitudes |f| and |g| are non-negative and measurable if f and g are (so it is ok to generalize as such.) Based on the conditions for equality to hold in Holder's Inequality, what are the conditions for equality to hold in Minkowski's Ineqaulity for complex f and g?