Exercise from An Introduction to Inequalities 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 ?
 A: 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.
A: 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.  
