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2006 Paper 2 Q6
D: 1600.0 B: 1516.0
vectors scalar product Cauchy-Schwarz inequality inequality equality condition proof

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: \[ (ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2) \,. \]

  1. Show that \((x+2y+2z)^2 \le 9(x^2+y^2+z^2)\) for all real numbers \(x\), \(y\) and \(z\).
  2. Find real numbers \(p\), \(q\) and \(r\) that satisfy both \[ p^2+4q^2+9r^2 = 729 \text{ and } 8p+8q+3r = 243\,. \]

Solution: 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 \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)

  1. By applying our inequality from the first part with \(a=1, b = 2, c=2\) we have \((x+2y+2z)^2 \leq (1+2^2+2^2)(x^2+y^2+z^2) = 9(x^2+y^2+z^2)\)
  2. 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 \[ (p,q,r) = \left (24, 6, 1 \right) \]

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