# A Cyclic Inequality in Three Variables with a Variable Hierarchy

### Solution 1

Recollecting a formula for the Vandermonde determinant,

\begin{align} &(y-x)(z-x)(z-y)\le 0&\Leftrightarrow\\ &\left|\begin{array}[ccc]\,1&1&1\\x&y&z\\x^2&y^2&z^2\end{array}\right|\le 0&\Leftrightarrow\\ &(xy^2+yz^2+zx^2)-(x^2y+y^2z+z^2x)\le 0&\Rightarrow\\ &2(x^2y+y^2z+z^2x)\ge xy^2+yz^2+zx^2+x^2y+y^2z+z^2x &\Leftrightarrow\\ &\small{2(x^2y+y^2z+z^2x+xyz)\ge xy^2+yz^2+zx^2+x^2y+y^2z+z^2x+2xyz}&\Leftrightarrow\\ &2(x^2y+y^2z+z^2x+xyz)\ge (x+y)(y+z)(z+x). \end{align}

### Solution 2

Let $y=z+p,x=y+q=z+p+q.\,$ We have

\displaystyle \begin{align} &2(x^2y+y^2z+z^2x+xyz)-(x+y)(y+z)(z+x)\\ &\qquad=xy(x-y)+yz(y-z)+zx(z-x)\\ &\qquad=(z+p+q)(z+p)q+(z+p)zp-z(z+p+q)(p+q)\\ &\qquad=pq(p+q)\ge 0. \end{align}

### Solution 3

Expanding $\displaystyle 2(x^2y + y^2z + z^2x+xyz) \ge (x+y) (y+z) (z+x)$ gives

$x^2y + y^2z + z^2x\ge xy^2 + yz^2 + zx^2,$

which is the same as $(x-y)(x-z)(y-z)\ge 0.$

As we saw, $xy^2+yz^2+zx^2\le x^2y+y^2z+z^2x.\,$ Then proceeding a little differently

\begin{align} &(xy^2+yz^2+zx^2)-(x^2y+y^2z+z^2x)\le 0&\Rightarrow\\ &2(xy^2+yz^2+zx^2)\le xy^2+yz^2+zx^2+x^2y+y^2z+z^2x &\Leftrightarrow\\ &\small{2(xy^2+yz^2+zx^2+xyz)\le xy^2+yz^2+zx^2+x^2y+y^2z+z^2x+2xyz}&\Leftrightarrow\\ &2(xy^2+yz^2+zx^2+xyz)\le (x+y)(y+z)(z+x). \end{align}

Thus we obtain a two sided inequality:

$\displaystyle 2\left(\sum_{cycl}xy^2+xyz\right)\le\prod_{cycl}(x+y)\le 2\left(\sum_{cycl}x^2y+xyz\right).$

### Acknowledgment

Leo Giugiuc has kindly posted the problem by Sladjan Stankovik at the CutTheKnotMath facebook page, along with an elegant solution of his (Solution 1). Leo also included a link to the original post by Seth Easterwood (Lorian Saceanu) at the mathematical inequalities facebook group.

Solution 2 is by Amit Itagi; Solution 3 is by N. N. Taleb.

• What Is Determinant?
• Determinant and Divisibility
• Determinants, Area, and Barycentric Coordinates
• Vandermonde matrix and determinant
• Cabart's Collinearity
• Three Similar Triangles
• Vandermonde Determinant
• Divisibility Of Prod-Dif And The Vandermonde Determinant
• Determinant on a Circle
• A Case for Determinants
•