Question

Prove that $||A+B||\mathbf{\le }||A||\mathbf{+}||B||$. This is called the triangle inequality; in two or three dimensions, it simply says that the length of one side of a triangle $\mathbf{\le }$ sum of the lengths of the other 2 sides. Hint: To prove it in n-dimensional space, write the square of the desired inequality using (10.2) and also use the Schwarz inequality (10.4). Generalize the theorem to complex Euclidean space by using (10.7) and (10.9).

Short answer

The triangle inequality,$||A+B||\le ||A||+||B||$

Step-by-step solution

Given information

$||A+B||\le ||A||+||B||$ is called triangle inequality in two or three dimensions. It says that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides.

Schwarz inequality

Let's take $\mathbf{<}\mathit{x}\mathbf{+}\mathit{y}\mathbf{,}\mathit{x}\mathbf{+}\mathit{y}\mathbf{>}\mathbf{=}{\left|\left|x+y\right|\right|}^{\mathbf{2}}$. Schwarz inequality states that for all vector $\mathit{x}$ and $\mathit{y}$, the inequality is written as $\left|<x,y>\right|\mathbf{\le }\mathbf{}\left|\left|x\right|\right|\mathbf{}\left|\left|y\right|\right|$.

Prove the triangle inequality

Take L.H.S., $\left|\left|A+B\right|\right|$

$$\begin{gathered} {\left|\left|A+B\right|\right|}^2=<A+B,A+B> \\ =<A,A>+<A,B>+<B,A>+<B,B> \\ ={\left|\left|A\right|\right|}^2+<A,B>+<\overline{A,B}>+{\left|\left|B\right|\right|}^2(\because <B,A>=<\overline{A,B}>) \\ ={||A||}^2+2Re<A,B>+{||B||}^2(\because <A,B>+<\overline{A,B}> \end{gathered}$$

Solve further.

$$\begin{gathered} =2Re⟨A,B⟩) \\ \le {||A||}^2+2\left|<A,B>\right|+{||B||}^2 \\ \le {||A||}^2+2||A||||B||+{||B||}^2 \\ ={(||A||+||B||)}^2 \end{gathered}$$

Thus, ${||A+B||}^2\le {(||A||+||B||)}^2.$

Take the square root for both sides .

$||A+B||\le ||A||+||B||$