Problem source

By considering a suitable scalar product, prove that 
\[
(ax+by+cz)^2 \le (a^2+b^2+c^2)(x^2+y^2+z^2)
\]
for any real numbers $a$, $b$, $c$, $x$, $y$ and $z$. Deduce a necessary and sufficient condition on  $a$, $b$, $c$, $x$, $y$ and $z$ for the following equation to hold:
\[
(ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2) \,.
\]
\begin{questionparts}
\item Show that $(x+2y+2z)^2 \le 9(x^2+y^2+z^2)$ for all real numbers $x$, $y$ and $z$.
\item Find real numbers $p$, $q$ and $r$ that satisfy both
\[
p^2+4q^2+9r^2 = 729
 \text{ and }
8p+8q+3r = 243\,.
\]
\end{questionparts}

Solution source

Consider $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$, $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$, then we know that

\begin{align*}
&& \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= \sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2} \cos \theta \\
\Rightarrow && (ax+by+cz)^2 &= (a^2+b^2+c^2)(x^2+y^2+z^2) \cos^2 \theta  \\
&&&\leq (a^2+b^2+c^2)(x^2+y^2+z^2) 
\end{align*}

For equality to hold, we must have that the vectors are parallel, ie  $\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}$

\begin{questionparts}
\item By applying our inequality from the first part with $a=1, b = 2, c=2$ we have $(x+2y+2z)^2 \leq (1+2^2+2^2)(x^2+y^2+z^2) = 9(x^2+y^2+z^2)$

\item Since 
\begin{align*}
&& (p^2+(2q)^2+(3r)^2)\left (8^2 +4^2+1^2 \right) &\geq (8p+8q+3r)^2 \\
\Leftrightarrow && 729 \cdot 81 &\geq 243^2 \\
&&3^6 \cdot 3^4 &\geq 3^{10}
\end{align*}
Therefore we must be in the equality case, ie $p = 8\lambda, 2q = 4\lambda, 3r = \lambda$ as well as $64\lambda + 16\lambda +\lambda = 243 \Rightarrow 81\lambda = 243 \Rightarrow \lambda = 3$ so we have

\[ (p,q,r) = \left (24, 6, 1  \right) \]

\end{questionparts}