# Is this combination of AM-GM and Muirhead valid?

I'm working on some inequality proofs involving cyclic and/or symmetric sums involving three positive variables $(a,b,c)$, and I found myself trying to use a plausible combination of Muirhead and AM-GM. But I have no proof that that combination is valid.

I will present the question for three variables, but the generalization to more than three variables is obvious.

Some notation: $\sigma$ will be used to represent all the permutations of $(a,b,c)$ so that $$\sum_\sigma a^p b^q c^r \equiv a^p b^q c^r + a^p c^q b^r + b^p c^q a^r +b^p a^q c^r + c^p a^q b^r + c^p b^q a^r$$ and we always take $p\geq q\geq r$, which we are free to do because of the symmetry of the sum.

Muirhead's inequality says (remember that $a,b,c$ are all positive)

If $[p_1 \geq p_2] \,\, \land \,\,[(p_1+q_1)\geq (p_2+q_2])\,\, \land \,\,[(p_1+q_1+r_1)\geq (p_2+q_2+r_2)]$ then $$\sum_\sigma a^{p_1} b^{q_1} c^{r_1} \geq \sum_\sigma a^{p_2} b^{q_2} c^{r_2}$$ I will word that if condition as "If $(p_1,q_1,r_1)$ majorizes $(p_2,q_2,r_2)$

In the applications I have, there are combinations of $\sum_\sigma a^p b^q c^r$ with various combinations of $p,q,r$ that always add up to the same number.

My conjecture is:

If $(p_1+q_1+r_1)=(p_2+q_2+r_2)=(p_3+q_3+r_3)$ and $(p_{12},q_{12},r_{12})$ majorizes $(p_3,q_3,r_3)$ where $p_{12}$ is the mean of $p_1$ and $p_2$ and similarly for $q_{12}, r_{12}$, then $$\frac{\sum_\sigma a^{p_1} b^{q_1} c^{r_1}+\sum_\sigma a^{p_2} b^{q_2} c^{r_2}}{2}\geq \sum_\sigma a^{p_3} b^{q_3} c^{r_3}$$

Note that by AM-GM $$\frac{\sum_\sigma a^{p_1} b^{q_1} c^{r_1}+\sum_\sigma a^{p_1} b^{q_1} c^{r_1}}{2}\geq \sqrt{\left( \sum_\sigma a^{p_1} b^{q_1} c^{r_1}\right) \left( \sum_\sigma a^{p_2} b^{q_2} c^{r_2}\right) }$$ but that is not good enough to prove the conjecture since it is not obvious that $\left( \sum_\sigma a^{p_1} b^{q_1} c^{r_1}\right) \left( \sum_\sigma a^{p_2} b^{q_2} c^{r_2}\right)$ is the same as $$\left( \sum_\sigma a^{p_{12}} b^{q_{12}} c^{r_{12}}\right)^2$$

Does anybody know how to prove this conjecture, or show that it is untrue?

• But isn't the last step (which you mention as non-obvious) just Cauchy Schwartz inequality? May 9, 2017 at 17:35
• Yes, @Macavity you are right. If you let the index $i$ in the C-S inequality range over the same order of permutations in both $\vec{u} = \sum a^{p_1/2} b^{q_1/2} c^{r_1/2}$ and $\vec{v} = \sum a^{p_2/2} b^{q_2/2} c^{r_2/2}$ then $$\left( \sum_\sigma a^{p_1} b^{q_1} c^{r_1}\right) \left( \sum_\sigma a^{p_2} b^{q_2} c^{r_2}\right) = (u\cdot v)^2 \geq |u|^2 |v|^2 = \left( \sum_\sigma a^{p_{12}} b^{q_{12}} c^{r_{12}}\right)^2$$ If you transfer this to be an answer, the question can be closed. May 9, 2017 at 18:34
• OK did that. Though how far this generalises seems an interesting question. May 9, 2017 at 18:46

As mentioned in comments, your argument can be completed using CS inequality as: $$\frac{\sum_\sigma a^{p_1} b^{q_1} c^{r_1}+\sum_\sigma a^{p_1} b^{q_1} c^{r_1}}{2}\geq \sqrt{\left( \sum_\sigma a^{p_1} b^{q_1} c^{r_1}\right) \left( \sum_\sigma a^{p_2} b^{q_2} c^{r_2}\right) } \geqslant \sum_{\sigma} a^{p_{12}}b^{q_{12}}c^{r_{12}}$$ which Muirhead assures is not lower than $\displaystyle \sum_\sigma a^{p_3} b^{q_3} c^{r_3}$.
More generally, using the same approach, if $(p_{12}, q_{12}, r_{12})$ are weighted averages of $(p_1, q_1, r_1)$ and $p_2, q_2, r_2)$, then using weighted AM-GM and Holder, it appears one can reach the same conclusion.