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2018 Paper 3 Q5
D: 1700.0 B: 1484.0
inequality AM-GM inequality sequences differentiation from first principles proof by induction polynomial analysis stationary points optimisation

The real numbers \(a_1\), \(a_2\), \(a_3\), \(\ldots\) are all positive. For each positive integer \(n\), \(A_n\) and \(G_n\) are defined by \[ A_n = \frac{a_1+a_2 + \cdots + a_n}n \ \ \ \ \ \text{and } \ \ \ \ \ G_n = \big( a_1a_2\cdots a_n\big) ^{1/n} \,. \]

  1. Show that, for any given positive integer \(k\), \[ (k+1) ( A_{k+1} - G_{k+1}) \ge k (A_k-G_k) \] if and only if \[\lambda^{k+1}_k -(k+1)\lambda_{{k}} +k \ge 0\,, \] where \( \lambda_{{k}} = \left(\dfrac{a_{k+1}}{G_{k}}\right)^{\frac1 {k+1}}\,\).
  2. Let \[ \f(x)=x^{k+1} -(k+1)x +k \,, \] where \(x > 0\) and \(k\) is a positive integer. Show that \(\f(x)\ge0\) and that \(\f(x)=0\) if and only if \(x = 1\,\).
  3. Deduce that:
    1. \(A_n \ge G_n\) for all \(n\); \\
    2. if \(A_n=G_n\) for some \(n\), then \(a_1=a_2 = \cdots = a_n\,\).

Solution:

  1. \begin{align*} && (k+1) (A_{k+1} - G_{k+1}) & \geq k(A_k - G_k) \\ \Leftrightarrow && \sum_{i=1}^{k+1} a_i - (k+1)G_{k+1} &\geq \sum_{i=1}^k a_i - kG_k \\ \Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} & \geq - k G_k \\ \Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} + k G_k& \geq 0\\ \Leftrightarrow && \frac{a_{k+1}}{G_k} -(k+1)G_k^{k/(k+1)-1}a_{k+1}^{1/(k+1)} + k & \geq 0\\ \Leftrightarrow && \lambda_k^{k+1} -(k+1)\lambda_k+ k & \geq 0\\ \end{align*} as required.
  2. \begin{align*} && f'(x) &= (k+1)x^k - (k+1) \\ &&&= (k+1)(x^k-1) \end{align*} Therefore \(f(x)\) is strictly decreasing on \((0,1)\) and strictly increasing on \((1,\infty)\) and so the minimum will be \(f(1) = 1 - (k+1) + k = 0\), so \(f(x) \geq 0\) with equality only at \(x = 1\).
    1. We can proceed by induction to show since the inequality holds for \(n=1\) and since if it holds for \(n=k\) it will hold for \(n=k+1\) as \(A_{k+1}-G_{k+1}\) must have the same sign as \(A_k - G_k\).
    2. The only way for equality to hold is if \(\lambda_k = 1\) for \(k = 1, \cdots n\), ie \(a_{k+1} = G_k\), but this means \(a_2 = a_1, a_3 = a_1\) etc. Therefore all values are equal.

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