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I'm working through some of the AM-GM problems in Larson's Problem-Solving Through Problems and I'm stuck on $7.2.11.b$. Part $a$ was proving the below

Let $x_i \in \mathbb{R}^+$ and $p_i \in \mathbb{Z}^+$, then prove

$$(x_1^{P_1}\cdots x_n^{P_n})^{1/p_1+\cdots+p_n} \leq \frac{p_1x_1 + \cdots + p_nx_n}{p_1 + \cdots p_n}.$$

That is a straightfoward application of AM-GM. Part $b$ however says to show this inequality holds when $p_i \in \mathbb{Q}^+$.

Let $p_i = \frac{a_i}{b_i}$ and $y_i = x_i^{1/b_i}$ then

$$(x_1^{P_1}\cdots x_n^{P_n})^{1/p_1+\cdots+p_n} = (y_1^{a_1}\cdots y_n^{a_n})^{1/p_1+\cdots+p_n}.$$

And since $p_1 + \cdots p_n \leq a_1 + \cdots a_n$, we can extend the inequality,

$$(y_1^{a_1}\cdots y_n^{a_n})^{1/p_1+\cdots+p_n} \leq (y_1^{a_1}\cdots y_n^{a_n})^{1/a_1+\cdots+a_n}.$$

Now we apply AM-GM and get that

$$(y_1^{a_1}\cdots y_n^{a_n})^{1/a_1+\cdots+a_n} \leq \frac{a_1y_1 + \cdots + a_ny_n}{a_1 + \cdots a_n} \leq \frac{a_1y_1 + \cdots + a_ny_n}{p_1 + \cdots p_n}.$$

And this is where I'm stuck. If $a_iy_i \leq p_ix_i$ then we're done. But I can't convince myself of that since $x$ can be in $(0,1)$. Can anyone nudge me in the right direction?

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1 Answer 1

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We may assume, by using a common denominator for all $p_i$'s, that $p_i = a_i/b$. Then we have \begin{align} (x_1^{p_1}\ldots x_n^{p_n})^{1/(p_1+\ldots p_n)} &= (x_1^{a_1/b}\ldots x_n^{a_n/b})^{1/(a_1/b + \ldots + a_n/b)} \\ &= (x_1^{a_1}\ldots x_n^{a_n})^{1/(a_1 + \ldots + a_n)} \\ &\leq \frac{a_1x_1 + \ldots a_nx_n}{a_1 + \ldots a_n}\\ &= \frac{(a_1/b)x_1 + \ldots (a_n/b)x_n}{a_1/b + \ldots a_n/b}\\ &= \frac{p_1x_1 + \ldots p_nx_n}{p_1 + \ldots p_n}, \end{align} where we have used the weighted AM-GM for positive integers in the middle line.

Old Answer:

If any of the $x_i$'s are $0$, we have equality. Otherwise, since the inequality is homogenous, it suffices to prove it for $\lambda x_i$, for some $\lambda$. Choose $\lambda$ large enough so that $(\lambda x_i)^{1/b_i} < \lambda x_i / b_i$ for all $x_i$. Then we have $a_i(\lambda x_i)^{1/b_i} \leq p_i (\lambda x_i)$.

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  • $\begingroup$ I don't know much about homogenous inequalities and the $\lambda$-trick but I'll read up on it. As for the choosing of $\lambda$, am I allowed to let that choice depend on $x$? If so, I could pick $\lambda_i > \left(\dfrac{b_i}{x_i^{b_i+1}}\right)^{1/b_i+1}$, then take $\lambda = \max{\lambda_i}$. $\endgroup$ Commented May 25, 2021 at 16:05
  • $\begingroup$ Homogenous means $f(\lambda x_1, ..., \lambda x_n) = \lambda^k f(x_1, ..., x_n)$. In the case of AM-GM, both sides are homogenous of degree $1$ ($k = 1$ above). So scaling all variables by a factor of $\lambda$ leaves the inequality unchanged. This means it suffices to show the inequality for your choice of scale. Regarding the choice of $\lambda$, yes, that is how you would do it. Another way of saying all this is that by homogeneity, we can assume without loss that all the $x_i$'s are at least $1$. $\endgroup$ Commented May 25, 2021 at 16:09
  • $\begingroup$ Thanks, that last sentence really puts it in perspective. $\endgroup$ Commented May 25, 2021 at 21:13
  • $\begingroup$ Actually, going over your proof, I think the inequality $(y_1^{a_1}\cdots y_n^{a_n})^{1/p_1+\cdots+p_n} \leq (y_1^{a_1}\cdots y_n^{a_n})^{1/a_1+\cdots+a_n}$ doesn't hold in general, since $1/(p_1+\ldots + p_n) \geq 1/(a_1+\ldots a_n)$. I've updated my answer to give a self contained proof. $\endgroup$ Commented May 25, 2021 at 22:59
  • $\begingroup$ Quite right about that inequality, that was an oversight from me. But I love your new proof. That feels more in line with the book than homogenous inequalities. Thank you again! $\endgroup$ Commented May 26, 2021 at 2:04

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