Question 1181369
Let {{{z[c]}}} denote the conjugate of the complex number {{{z}}}.


Then {{{abs(z[1]+z[2])^2   = (z[1] + z[2])[c]*(z[1]+z[2])  }}}


= {{{(z[1c] + z[2c])*(z[1]+z[2]) = z[1c]*z[1] + z[1c]*z[2] + z[2c]*z[1] + z[2c]*z[2]  = z[1c]*z[1] + (z[1c]*z[2] + z[2c]*z[1]) + z[2c]*z[2]  }}}


= {{{abs(z[1])^2 + 2*Re(z[1c]*z[2]) + abs(z[2])^2}}}, from the fact Re(z) = (z + conjugate (z)) / 2.


{{{abs(z[1])^2 + 2*Re(z[1c]*z[2]) + abs(z[2])^2 <= abs(z[1])^2 +2*abs(z[1c]*z[2]) +abs(z[2])^2= abs(z[1])^2 +2*abs(z[1c])*abs(z[2]) +abs(z[2])^2

= abs(z[1])^2 +2*abs(z[1])*abs(z[2]) +abs(z[2])^2
}}}, from the fact  Re(z) < or = |z|.


===> {{{abs(z[1]+z[2])^2 <= abs(z[1])^2 +2*abs(z[1])*abs(z[2]) +abs(z[2])^2

= (abs(z[1])+abs(z[2]))^2}}}

Since both {{{abs(z[1]+z[2])}}} and {{{abs(z[1])+abs(z[2])}}} are non-negative, the last inequality implies that 


{{{abs(z[1]+z[2]) <= abs(z[1])+abs(z[2])}}}, and the Triangle Inequality is proved.