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QUESTION Let $\textbf{u}\neq \textbf{0}$ and
$\textbf{v}\in\textbf{R}^n$. Consider the vector $\lambda
\textbf{u+v}$ where $\lambda \in \textbf{R}$, then by positivity

$0\leq(\lambda\textbf{u+v}).(\lambda\textbf{u+v})=(\textbf{u.u})
\lambda^2+2(\textbf{u.v})\lambda+(\textbf{v.v})$.

\begin{description}

\item[(a)]
Considered as a polynomial in $\lambda$, how many real roots does

$p(\lambda)=(\textbf{u.u}
\lambda^2+2(\textbf{u.v})\lambda+(\textbf{v.v})$

have?

\item[(b)]

Write down the discriminant of the polynomial $p(\lambda)$.

\item[(c)]

What does the result in part (a) tell you about the discriminant?

\item[(d)]

Deduce the Cauchy-Schwarz inequality: \ \
$(\textbf{u.v})^2\leq(\textbf{u.u})(\textbf{v.v})$.

\item[(e)]

Prove the triangle inequality in $\textbf{R}^n:\ \
|\textbf{u+v}|\leq|\textbf{u}|+|\textbf{v}|$

[Hint: Rewrite in terms of the inner product:
$|u|=\sqrt{(\textbf{u.u})}$, etc.]

\item[(f)]

Prove that in $\textbf{R}^n: \ \ \
|\textbf{u+v}|^2+|\textbf{u-v}|^2=2(|\textbf{u}|^2+|\textbf{v}|^2)$.

\end{description}


ANSWER
\begin{description}

\item[(a)]
Since $p(\lambda)\geq0$ for all $\lambda$, there is at most one
real root.

\item[(b)]
$\Delta=4(\textbf{u.v})^2-4(\textbf{u.u})(\textbf{v.v}).$

\item[(c)]
Since $p(\lambda)$ has at most one real root, $\Delta \leq 0$.

\item[(d)]
Divide $\Delta\leq0$ by 4.

\item[(e)]
In terms of inner products the desired result can be written

\begin{tabular}{cl}
&amp;$\sqrt{\{(\textbf{u+v}).(\textbf{u+v})\}}\leq\sqrt{(\textbf{u.u})}+\sqrt{(\textbf{v.v})}$\\
$\Leftrightarrow$&amp;$(\textbf{u+v}).(\textbf{u+v})\leq
\textbf{u.u}+2\sqrt{\{(\textbf{u.u})(\textbf{v.v})\}}+(\textbf{v.v})$\\
$\Leftrightarrow$&amp;$\textbf{u.u}+2\textbf{u.v}+\textbf{v.v}\leq
\textbf{u.u}+2\sqrt{\{(\textbf{u.u})(\textbf{v.v})\}}+(\textbf{v.v})$\\
$\Leftrightarrow$&amp;$\textbf{u.v}\leq\sqrt{\{(\textbf{u.u})(\textbf{v.v})\}}$\\
$\Leftarrow$&amp;$(\textbf{u.v})^2\leq(\textbf{u.u})(\textbf{v.v})$
\end{tabular}

but this is just the Cauchy-Schwarz inequality.

\item[(f)]
\begin{eqnarray*}
|\textbf{u+v}|^2+|\textbf{u-v}|^2&amp;=&amp;(\textbf{u+v}).(\textbf{u+v})+(\textbf{u-v}).(\textbf{u-v})\\
&amp;=&amp;\textbf{u.u}+2\textbf{u.v}+\textbf{v.v}+\textbf{u.u}-2\textbf{u.v}+\textbf{v.v}\\
&amp;=&amp;2(\textbf{u.u}+\textbf{v.v})\\
&amp;=&amp;2(|\textbf{u}|^2+|\textbf{v}|^2).
\end{eqnarray*}

\end{description}



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