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Section 1.5 Vector Inequalities
Theorem 1.5.1 . Cauchy-Schwarz Inequality.
For any two vectors \(\vec{u}\) and \(\vec{v}\text{,}\) we have
\begin{equation*}
\left| \vec{u} \cdot \vec{v} \right| \le \Vert \vec{u} \Vert \Vert \vec{v} \Vert,
\end{equation*}
with equality if and only if \(\vec{u}\) is a scalar multiple of \(\vec{v}, \) or one of them is the zero vector.
Proof.
By definition,
\begin{equation*}
\vec{u} \cdot \vec{v} = \Vert \vec{u} \Vert \Vert \vec{v} \Vert \cos \theta,
\end{equation*}
where \(\theta\) is the angle between the two vectors. Taking absolute values on both sides gives,
\begin{equation*}
\begin{array}{ccc}
\left| \vec{u} \cdot \vec{v} \right| \amp =\amp \left|\Vert \vec{u} \Vert \Vert \vec{v} \Vert \cos \theta\right|,\\
\amp = \amp \Vert \vec{u} \Vert \Vert \vec{v} \Vert \left|\cos \theta\right|,\\
\amp \le \amp \Vert \vec{u} \Vert \Vert \vec{v} \Vert.
\end{array}
\end{equation*}
Since \(\left|\cos \theta\right| \le 1\) for any angle \(\theta\) \(\qed\)
\(~\)
Example 1.5.2 . Cauchy-Schwarz inequality.
Verify the Cauchy-Schwarz inequality for \(\vec{u} = -\hi + \hj + \hk\) and \(\vec{v} = 2\hi + \hj - 3\hk\text{.}\)
\begin{equation*}
\begin{array}{ccc}
\vec{u} \cdot \vec{v} \amp = \amp -2 +1 -3 = -4\\
\left|\vec{u} \cdot \vec{v}\right| \amp = \amp 4\\
\amp \amp\\
\Vert \vec{u} \Vert \amp =\amp \sqrt{(-1)^2 + 1 + 1} = \sqrt{3}\\
\Vert \vec{v} \Vert \amp =\amp \sqrt{2^2 + 1 + (-3)^2} = \sqrt{14}\\
\Vert \vec{u} \Vert\Vert \vec{v} \Vert \amp = \amp \sqrt{3} \cdot \sqrt{14} = \sqrt{42} \approx 6.481.\\
\amp \Rightarrow \amp\\
\left|\vec{u} \cdot \vec{v}\right| \amp \lt \amp \Vert \vec{u} \Vert\Vert \vec{v} \Vert.\\
(4) \amp \amp (6.481)
\end{array}
\end{equation*}
\(~\)
Theorem 1.5.3 . Triangle Inequality.
For any two vectors \(\vec{u}\) and \(\vec{v}\text{,}\) the following is always true
\begin{equation*}
\Vert \vec{u} + \vec{v} \Vert \le \Vert \vec{u} \Vert + \Vert \vec{v} \Vert,
\end{equation*}
with equality if and only one of the vectors is the zero vector.
Proof.
\begin{equation*}
\begin{array}{ccc}
\Vert \vec{u} + \vec{v} \Vert^2 \amp =\amp \left( \vec{u} + \vec{v} \right) \cdot
\left( \vec{u} + \vec{v} \right)\\
\amp = \amp \vec{u} \cdot \vec{u} + 2\, \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{v}\\
\amp = \amp \Vert \vec{u} \Vert^2 + 2\, \vec{u} \cdot \vec{v} + \Vert \vec{v} \Vert^2
\end{array}
\end{equation*}
We use the Cauchy-Schwarz inequality to express the dot product, in the middle, in terms of the magnitudes,
\begin{equation*}
\begin{array}{ccc}
\Vert \vec{u} \Vert^2 + 2\, \vec{u} \cdot \vec{v} + \Vert \vec{v} \Vert^2
\amp \le \amp \Vert \vec{u} \Vert^2 + 2\, \Vert \vec{u} \Vert \Vert \vec{v} \Vert + \Vert \vec{v} \Vert^2\\
\amp \le \amp \left(\Vert \vec{u} \Vert + \Vert \vec{v} \Vert \right)^2.
\end{array}
\end{equation*}
Thus,
\begin{equation*}
\Vert \vec{u} + \vec{v} \Vert^2 \le \left(\Vert \vec{u} \Vert + \Vert \vec{v} \Vert \right)^2.
\end{equation*}
Taking the roots on both sides proves the theorem \(\qed\)
\(~\)
Example 1.5.4 . Triangle inequality.
Verify that unlike the Cauchy-Schwarz inequality; the equality in the triangle inequality does not hold when one of the vectors is a scalar multiple of the other.
\begin{equation*}
\begin{array}{ccc}
\vec{u} \amp = \amp 3\hi - 2\hj + \hk\\
\vec{v} \amp = \amp -6\hi + 4\hj - 2 \hk\\
\amp \amp \\
\Vert \vec{u} \Vert \amp = \amp = \sqrt{9+4+1} = \sqrt{14}\\
\Vert \vec{v} \Vert \amp = \amp = \sqrt{36+16+4} = \sqrt{56} = 2 \sqrt{14}\\
\Vert \vec{u}\Vert + \Vert \vec{v} \Vert \amp = \amp 3\sqrt{14}.\\
\amp \amp \\
\vec{u} + \vec{v} \amp = \amp -3 \hi + 2\hj -1\hk\\
\Vert \vec{u} + \vec{v} \Vert \amp = \amp \sqrt{(-3)^2 + 2^2 + (-1)^2} = \sqrt{14}\\
\end{array}
\end{equation*}
\begin{equation*}
\begin{array}{ccc}
\amp \amp \\
\Vert \vec{u} + \vec{v} \Vert \amp = \amp \displaystyle \frac{1}{2} \Vert \vec{u}\Vert + \Vert \vec{v} \Vert \\
\amp \Rightarrow \amp \\
\Vert \vec{u} + \vec{v} \Vert \amp \ne \amp \Vert \vec{u}\Vert + \Vert \vec{v} \Vert. \\
\end{array}
\end{equation*}
