Tingle Inequality

Statement:- In an inner product space over F then prove that ||α+β||≤||α||+||β||.

Proof:- Since, ||α+β||=√(α+β,α+β)         

⇨                    ||α+β||²=(α,β,α,β)

⇨                     ||α+β||²=(α,α+β)+(β,α+β) {by linearity}

=(α,α)+(α,β)+(β,α)+(β,β)

=||α||²+(α,β)+(β,α)+||β||²

=||α||²+(α,β)+( )+||β||²

{since(α,α)=||α||²&(β,α)=( )

=||α||²+2Re(α,β)+||β||²

since , Re(α,β)≤|(α,β)| {since +Z=2x=2Re(z)}

⇨    ||α||²+2Re(α,β)+||β||²≤||α||²+2|(α,β)|+||β||²

||α+β||²≤||α||²+2|(α,β)|+||β||²

≤||α||²+2||α||||β||+||β||²

(since |(α,β)|≤||α||||β||

⇨                                 ||α+β||²≤(||α||+||β||)²

taking square root

⇨                                  ||α+β||≤||α||+||β||      PROVED.