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I am reading An Introduction to Inequalities by Beckenbach and Bellman and on chapter 4.3 there is this exercise. It's regarding AM-GM inequality.

How can I prove it? I can't figure it out.

$$ {m_1y_1+m_2y_2+\cdots+m_ky_k \over m_1+m_2+\cdots+m_k} \ge \sqrt[\large m_1+m_2+\cdots+m_k]{y_1^{m_1}\cdot y_2^{m_2}\cdots y_k^{m_k}} $$

The authors has this hint (consider AM-GM written using $a_1, a_2,\dots$):

In the arithmetic-mean-geometric-mean inequality (4. 19) set the first $m_1$ of the numbers $a_i$ equal to the same value $y_1$, set the next $m_2$ of the numbers $a_i$ equal to the same number $y_2$, and set the last $m_k$ of the numbers $a_i$ equal to the same value $y_k$, and observe that $m_1+m_2+\dots+m_k=n$.

But I just don't understand what he means by this. Substitute what with what ?

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  • $\begingroup$ A good question, but it would be even better if you shared some of your ideas concerning it. What have you tried so far? $\endgroup$ – Daniel W. Farlow Mar 24 '15 at 18:13
  • $\begingroup$ @crash That's the problem. I don't know how to attack it. I've tried figuring out how to use AM-GM in substituting something like a1 = m1y1, a2=m2y2 but I can't figure it out. $\endgroup$ – amb Mar 24 '15 at 18:17
  • $\begingroup$ Then I would put that in the question. $\endgroup$ – Daniel W. Farlow Mar 24 '15 at 18:21
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In that text, you start with the AM-GM: $$\frac{a_1+a_2+\dots+a_n}n \ge \sqrt[n]{a_1 a_2 \dots a_n} \tag{4.19}$$

and you need to show for positive integers $m_i$, and non-negative reals $y_i$ we have $${m_1y_1+m_2y_2+\cdots+m_ky_k \over m_1+m_2+\cdots+m_k} \ge \sqrt[\large m_1+m_2+\cdots+m_k]{y_1^{m_1}\cdot y_2^{m_2}\cdots y_k^{m_k}}$$

As $(4.19)$ holds for any positive integer $n$ and any non-negative $a_i$, we may set the $n=m_1+m_2+\dots+m_k$, and set the first $m_1$ values for $a_i$ to be $y_1$ etc. This gives the modified $(4.19)$ as

$$\frac{\overbrace{(y_1+y_1+\dots+y_1)}^{m_1 \text{ times}}+\overbrace{(y_2+y_2+\dots+y_2)}^{m_2 \text{ times}}+\dots+\overbrace{(y_k+y_k+\dots+y_k)}^{m_k \text{ times}}}n \ge \sqrt[n]{\overbrace{(y_1 \cdot y_1 \dots y_1)}^{m_1 \text{ times}}\, \overbrace{(y_2 \cdot y_2 \dots y_2)}^{m_2 \text{ times}} \dots \overbrace{(y_k \cdot y_k \dots y_k)}^{m_k \text{ times}}} $$

which is of course the same as: $$\iff \frac{m_1y_1+m_2y_2+\dots+m_k y_k}n \ge \sqrt[n]{y_1^{m_1} y_2^{m_2} \dots y_k^{m_k}} $$

Using $n=m_1+m_2+\dots+m_k$ completes the proof.

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  • $\begingroup$ So the idea that I had, in the comment of the question, of using $a_i=m_iy_i$ was useful... But what I wasn't sure about is just setting $n=m_1+...+m_k$. I had the feeling it would restrict the proof so it would not be quite right. $\endgroup$ – amb Mar 24 '15 at 20:04
  • $\begingroup$ The idea that I had was actually not right. $\endgroup$ – amb Mar 24 '15 at 20:16
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hint: Consider the function

$$f(y_1,y_2,\cdots, y_k) = \ln(a_1y_1+a_2y_2+\cdots + a_ky_k), a_1+a_2+\cdots + a_k = 1, a_k > 0$$, and use Jensen's inequality.

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  • $\begingroup$ Thank you for the answer, but I have accepted another answer that was inline with what the authors meant. $\endgroup$ – amb Mar 24 '15 at 20:18

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