3

Let's define:

$$M[a, b, c]=\sum_{cyc}x^a y^b z^c $$

I need to prove that for all positive and real $x$, $y$ and $z$:

$$M[6, 3, 0] + M[3, 3, 3] \ge M[5, 2, 2] + M[4, 4, 1]$$

From Muirhead's inequality it's obvious that $M[6,3,0]$ is the biggest. But $M[3,3,3]$ is the smallest cyclic sum. So it looks like that it's not possible prove the inequality by using Muirhead alone.

Schur's inequality also did not help me much: $M[a+2b,0,0] + M[a,b,b] \ge 2M[a+b,b,0]$

Is there any general method how to approach problems like this one?

Michael Rozenberg
  • 194,933
Saša
  • 15,906
  • $\sum_{cyc}$ actually stands for $\sum_{sym}$. In other words $\sum_{cyc}x^2 = 2x^2 + 2y^2 + 2z^2$ – Saša Apr 03 '18 at 20:04

1 Answers1

3

Let $\frac{a}{c}=x$, $\frac{b}{a}=y$ and $\frac{c}{b}=z$.

Hence, $xyz=1$ and after dividing of the both sides by $a^3b^3c^3$ we need to prove that $$\sum_{cyc}\left(x^3-\frac{x}{y}-\frac{y}{x}+1\right)\geq0$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.

Michael Rozenberg
  • 194,933