**Let \(m_1, m_2 , … , m_k \) be k positive numbers such that their reciprocals are in A.P. Show that \(k< m_1 + 2 \) . Also find such a sequence for positive integer k > 0.**

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*Teacher:* This is a simple application of Arithmetic Mean – Harmonic Mean Inequality. First notice that if \(k \le 2 \) we have nothing to prove as \(m_1 \ge 1\) , \(m_1 +2 \) is definitely greater than k.

So we are interested in the cases where \(k \ge 3 \)

Now apply the A.M. – H.M. inequality

*Student:* Ok. \(\displaystyle { \frac {k}{\frac{1}{m_1} + \frac {1}{m_2} + … + \frac{1}{m_k}} \ge \frac{m_1 + m_2 + … + m_k}{k} } \)

Now we already know that \(\displaystyle { \frac{1}{m_1} , \frac {1}{m_2} , … , \frac{1}{m_k} } \) are in A.P. So sum of these k terms in A.P. is (first term plus last term) times number of terms by 2. In other words \(\displaystyle { \frac{1}{m_1} + \frac {1}{m_2} + … + \frac{1}{m_k} = \frac {k}{2} \left ( \frac{1}{m_1} + \frac{1}{m_k} \right)} \)

So our expression becomes \(\displaystyle { \frac {k}{ \frac {k}{2} \left ( \frac{1}{m_1} + \frac{1}{m_k} \right)} \le \frac{m_1 + m_2 + … + m_k}{k} } \)

This implies \(\displaystyle { k \le (m_1 + m_2 + … + m_k) \frac {\left ( \frac{1}{m_1} + \frac{1}{m_k} \right)}{2} } \)

I cannot think what to do next.

*Teacher:*** **It is quite simple actually. Notice that \(m_1 , … , m_k \) are integers then their reciprocals are less than (or at most equal to) 1.

Hence \(\displaystyle { \frac {1}{m_1} \le 1 , \frac{1}{m_k} \le 1 \Rightarrow \frac{1}{m_1} + \frac{1} {m_k} \le 2 }\). \(m_1 \) and \(m_k \) are different hence both cannot be 1.

Thus \(\displaystyle { \frac{1}{m_1} + \frac{1} {m_k} < 2 \Rightarrow \frac {\left ( \frac{1}{m_1} + \frac{1}{m_k} \right)}{2} < 1 }\)

Hence \(\displaystyle { k \le (m_1 + m_2 + … + m_k) \frac {\left ( \frac{1}{m_1} + \frac{1}{m_k}\right)}{2} < (m_1 + m_2 + … + m_k) \le m_1 + k-1} \) since \(1 \le m_2 , … , 1 \le m_k \)

Now k is at least 3 (we are already done with k < 3). So \(k < m_1 + k-1 \le m_1 + 2 \)