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2008 Paper 1 Q3
D: 1500.0 B: 1484.0
proof inequality algebraic manipulation rearrangement inequality AM-GM inequality symmetric functions WLOG ordering

Prove that, if \(c\ge a\) and \(d\ge b\), then \[ ab+cd\ge bc+ad\,. \tag{\(*\)} \]

  1. If \(x\ge y\), use \((*)\) to show that \(x^2+y^2\ge 2xy\,\). If, further, \(x\ge z\) and \(y\ge z\), use \((*)\) to show that \(z^2+xy\ge xz+yz\) and deduce that \(x^2+y^2+z^2\ge xy+yz+zx\,\). Prove that the inequality \(x^2+y^2+z^2\ge xy+yz+zx\,\) holds for all \(x\), \(y\) and \(z\).
  2. Show similarly that the inequality \[\frac st +\frac tr +\frac rs +\frac ts +\frac rt +\frac sr \ge 6\] holds for all positive \(r\), \(s\) and \(t\).

Solution: \begin{align*} && \underbrace{(c-a)}_{\geq 0}\underbrace{(d-b)}_{\geq 0} & \geq 0 \\ \Leftrightarrow && cd -bc -ad + ab &\geq 0 \\ \Leftrightarrow && ab +cd &\geq bc+ad \\ \end{align*}

  1. Applying \((*)\) with \(c=d=x\) and \(a=b=y\) we obtain: \(x^2 + y^2 \geq xy + xy = 2xy\) Similarly, applying \((*)\) with \(c=x, d=y, a=b=z\) we obtain: \(z^2 + xy \geq zx+zy\) so \(x^2+y^2+z^2 \geq 2xy + z^2 \geq xy + zx+zy\) There was nothing special about our choice of ordering \(x,y,z\) so it is true for all \(x,y,z\)
  2. \begin{align*} \frac st +\frac tr +\frac rs +\frac ts +\frac rt +\frac sr &=\left ( \frac st+\frac ts \right)+\left ( \frac tr +\frac rt \right)+\left ( \frac rs +\frac sr \right) \\ & \geq 2 \sqrt{\frac st \frac ts} + 2 \sqrt{\frac tr \frac rt} + 2 \sqrt{\frac rs \frac sr} \\ & = 2 + 2 + 2 \\ &= 6 \end{align*}

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