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 ?

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.