The proof follows from squaring both sides and applying one of the intermediate steps from our [proof](Cauchy-Schwarz%20inequality.md#Proof%20of%20the%20Cauchy-Schwarz%20inequality%20on%20complex%20vector%20space%20Complex%2020vector%2020spaces%20md%20s) of the [[Cauchy-Schwarz inequality]] on [complex vector spaces:](Complex%20vector%20spaces.md) $(||u+v||)^2 = \langle u+v, u+v\rangle = \langle u,u\rangle + \langle v,v\rangle + \langle u,v\rangle + \langle v,u\rangle$ $=||f||^2 + ||v||^2 + 2\mbox{Re}{(\langle u,v\rangle)}$ ^01992a We find that: $||u||^2 + ||v||^2 + 2\mbox{Re}{(\langle u,v\rangle)} \leq ||u||^2 + ||v||^2 + 2||u||\,||v|| = (||u||+||v||)^2$ since we know from the [proof](Cauchy-Schwarz%20inequality.md#Proof%20of%20the%20Cauchy-Schwarz%20inequality%20on%20complex%20vector%20space%20Complex%2020vector%2020spaces%20md%20s) of the [[Cauchy-Schwarz inequality]] on [complex vector spaces](Complex%20vector%20spaces.md) that [$\mbox{Re}{(\langle u, v \rangle)} \leq ||u||||v||$](proof%20of%20the%20Cauchy-Schwarz%20inequality%20on%20complex%20vector%20spaces#^fd6dd8) we can deduce that $||u+v|| \leq ||u|| + ||v||$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\blacksquare$ ^e64d56 #MathematicalFoundations/Algebra/AbstractAlgebra/LinearAlgebra #MathematicalFoundations/Analysis/FunctionalAnalysis #Proofs