Conditions for Muirhead's inequality hold for cyclic sums

Question

I know that Muirhead's inequality apply only for symmetrical sums, but all inequalities with cyclic sums I have seen have the sequence in the greater side majorizing the sequence in the smaller side (what is necessary for Muirhead's inequality hold for symmetrical sums). Also I couldn't find any counter-example that shows Muirhead's inequality doesn't hold for cyclic sums, so I wonder if the conditions for Muirhead's inequality hold in a cyclic sum are known, because if they are known, and easier to check than to prove a "hard" inequality then it would make way easier to prove a lot of other inequalities.

Answer 1

For a cyclic sum the majorization is not enough.

For example, for non-negative variables $(3,2,0)\succ(3,1,1)$ but the inequality $$a^3b^2+b^3c^2+c^3a^2\geq a^3bc+b^3ac+c^3ab$$ is wrong.

Try $a\rightarrow+\infty$ and $b^2-bc<0$.

By the way, there is the following way to prove of the Murhead's type cyclic inequalities.

We'll prove that $$\sum_{cyc}a^5b^2\geq \sum_{cyc}a^4b^2c$$ for non-negative variables.

Indeed, by AM-GM $$\sum_{cyc}a^5b^2=\frac{1}{19}\sum_{cyc}(14a^5b^2+2b^5c^2+3c^5a^2)\geq\sum_{cyc}\sqrt[19]{\left(a^5b^2\right)^{14}\left(b^5c^2\right)^2\left(c^5a^2\right)^3}=\sum_{cyc}a^4b^2c.$$ Vasile Cirtoaje was first, which proved that if this way does not work, so the inequality is wrong.

For example, we'll try to prove that $$\sum_{cyc}a^3b^2\geq \sum_{cyc}a^3bc$$ by this way.

We'll try to find values of $\alpha$, $\beta$ and $\gamma$ such that $\alpha+\beta+\gamma=1$ and the inequality $$\alpha a^3b^2+\beta b^3c^2+\gamma c^3a^2\geq a^3bc$$ would be true by AM-GM.

Indeed, by AM-GM $$\alpha a^3b^2+\beta b^3c^2+\gamma c^3a^2\geq a^{3\alpha+2\gamma}b^{2\alpha+3\beta}c^{2\beta+3\gamma}$$ and we obtain the following system:

$3\alpha+2\gamma=3,$ $2\alpha+3\beta=1$ and $\alpha+\beta+\gamma=1$, which gives $$(\alpha,\beta,\gamma)=\left(\frac{5}{7},-\frac{1}{7},\frac{3}{7}\right)$$ and since $-\frac{1}{7}<0$, this way does not give a proof, which says that the inequality is wrong.

Answer 2

Here's a Necessary and Sufficient condition for a true $3-$ variebles cyclic Muirhead-like Inequality. (Assuming all the variables are positive)

(I strongly suggest that you only read the 'Theorem' and 'Example' parts, and if you wish to learn more about ToC Method, please visit the link below)

Theorem. Consider the 'Triangle of Coefficient' (ToC for brevity) of the inquality you wish to prove, a $3-$ variebles cyclic Muirhead-like Inequality is true if and only if all the $-1$ is within the convex hull constructed by the $1$ . (In this case this convex hull is always a triangle, and it's also true if $-1$ lands on the side. Also note that due to cyclic symmetry, the ToC is rotationally symmetric by $2\pi/3$ in terms of its centroid.) $\square$

The proof for neccesity is quite simple because this determination works for any inequality, thus I won't prove it here and I'll focus on the proof for sufficiency.

Hint for the proof for neccessity: by AM-GM, we get the feeling of that the more unevenly-distrubuted the exponents are, the stronger the term is.

Here I'll first use the neccessity part of this theorem to cope with a counterexample, you may gain a basic idea of ToC Method as well.

Example1. Let's focus on a specific counterexample which Michael Rozenberg has provided.

Let $f:=\sum{a^3b^2}-\sum{a^3bc}$ , its ToC

$$ \begin{bmatrix} 0&&0&&1&&0&&0&&0\\ &0&&-1&&0&&-1&&0\\ &&0&&0&&0&&1\\ &&&1&&-1&&0\\ &&&&0&&0\\ &&&&&0 \end{bmatrix} $$

Simply draw the triangle constructed by $1$, its' clear that the $-1$ are all outside of this triangle , which immediately falsifies this inequality. We may construct many other counterexamples by using this determination backwards. $\square$

Finally I'll give a proof for sufficiency. (This proof is extremely not rigorous because there's not yet a set of terms for ToC method. I just hope it conveys the basic ideas. If you can't stand this proof (I can't myself) , please just skip to the next part, in which I'll demonstrate the idea of this proof with an example.)

Proof. Note that the ToC of any $3-$ variebles cyclic Muirhead-like Inequality consists of three elements $1$ and three elements $-1$ . Consider a specific $-1$ and denote its position in the ToC as $P$ , and the three $1$ as $A,B,C$ .

Now that $P$ is inside the triangle $A,B,C$ , let $a,b,c$ respectively represent $S_{\triangle{BPC}},S_{\triangle{APC}},S_{\triangle{CPA}}$ .

Because any binary AM-GM results in a pair of affined unit $[1,-2,1]$ , we get the insight that AM-GM is associated with midpoints, which corresponds to centroids in trenary cases. By Lami's Theorem, the ratio of the coefficients in AM-GM equals to the ratio $a:b:c$ . By writing down this AM-GM, we have completed the proof. $\square$

Example2. For positive real numbers $a,b,c$, prove that:

$$ a^3b+b^3c+c^3a\geqslant abc(a+b+c) $$

Let $g:=\sum{a^3b}-\sum{a^2bc}$ , its ToC

$$ \begin{bmatrix} 0&&1&&0&&0&&0\\ &0&&-1&&-1&&1\\ &&0&&-1&&0\\ &&&1&&0\\ &&&&0 \end{bmatrix} $$

By applying Pick's Theorem, we may quickly calculate that the ratio of coefficients is $1:2:4$ , which implies this AM-GM Inequality

$$ \sum(a^3b+2b^3c+4c^3a)\geqslant\sum{7\left(a^7b^7c^14\right)^{\tfrac17}}=7\sum{abc^2} $$

Simply by rearranging the terms proves the desired inequality. $\square$

Example3. This is an example Michael Rozenberg used in his answer, with ToC it's easier to figure out the coefficients.

For positive real numbers $a,b,c$, prove that:

$$ a^5b^2+b^5c^2+c^5a^2\geqslant abc(a^3b+b^3c+c^3a) $$

Let $h:=\sum{a^5b^2}-\sum{a^4b^2c}$ , its ToC

$$ \begin{bmatrix} 0&&0&&1&&0&&0&&0&&0&&0\\ &0&&0&&-1&&0&&0&&0&&0\\ &&0&&0&&0&&0&&-1&&1\\ &&&0&&0&&0&&0&&0\\ &&&&0&&-1&&0&&0\\ &&&&&1&&0&&0\\ &&&&&&0&&0\\ &&&&&&&0 \end{bmatrix} $$

By applying Pick's Theorem, we may quickly calculate that the ratio of coefficients is $2:3:14$ , which implies this AM-GM Inequality

$$ \sum(2a^5b^2+3b^5c^2+14c^5a^2)\geqslant\sum19\left(a^{38}b^{19}c^{76}\right)^{\tfrac1{19}}=19\sum{a^2bc^4} $$

Simply by rearranging the terms proves the desired inequality. $\square$

You may see from Example2 and Example3 that this method is rather stereotyped, and I belive you could even write a programme specialised to solve such problems. This actually suggests the validity of this theorem.