# Cyclic Inequality In Three Variables from the 2018 Romanian Olympiad, Grade 9

### Solution 1

Multiplying out and using the constraint to homogenize the inequality we reduce it to

$(a+b+c)^3+9(ab^2+bc^2+ca^2)-6(a+b+c)(ab+bc+ca)\ge 0,$

which is a homogeneous inequality with three variables of degree three and, therefore, submits to the CD3 theorem by Pham-Citroaje-Zhou theorem which stipulates that, for an inequality $f(x,y,z)\ge 0$ to hold (where $f$ is as just described) suffice it to verify that $f(1,1,1)\ge 0$ and that, for all $x,y\ge 0,$ $f(x,y,0)\ge 0.$ As the inequality is cyclic, we can as well check whether $f(y,0,x)\ge 0;$ and, since the inequality is homogeneous, suffice it to check that $f(y,0,1)\ge 0.$ The verification is immediate:

$\begin{cases} (1,1,1):&3^3+9\cdot 3-6\cdot 3\cdot 3=0\\ (y,0,1):&y^3+(6y^2-3y+1)\ge 0. \end{cases}$

Equality is obviously attained for $a=b=c=1.$

### Solution 2

We have $0\le a,b,c\le 3.$ Let $x=a-1,$ $y=b-1,$ $z=c-1.$ Then $-1\le,x,y,z\le 2$ and $x+y+z=0.$ The inequlity becomes

\displaystyle\begin{align} &\frac{x}{y+2}+\frac{y}{z+2}+\frac{z}{x+2}\ge 0\;&\Leftrightarrow\\ &-2(xy+yz+zx)+xy^2+yz^2+zx^2\ge 0\;&\Leftrightarrow\\ &x^2+y^2+z^2+xy^2+yz^2+zx^2\ge 0\;&\Leftrightarrow\\ &(x+1)y^2+(y+1)z^2+(z+1)x^2\ge 0, \end{align}

which is obviouis.

### Solution 3

After multiplying out and simplifying, the required inequality is equivalent to

\displaystyle \begin{align} &\sum_{cycl}ab^2+\sum_{cycl}a^2\ge 6\;&\Leftrightarrow\\ &\sum_{cycl}ab^2+\left(\sum_{cycl}a\right)^2-2\sum_{cycl}ab\ge 6\;&\Leftrightarrow\\ &\sum_{cycl}ab^2-2\sum_{cycl}ab\ge -3\;&\Leftrightarrow\\ &\sum_{cycl}ab^2-2\sum_{cycl}ab+\sum_{cycl}a\ge -3+\sum_{cycl}a\;&\Leftrightarrow\\ &\sum_{cycl}a(b-1)^2\ge 0. \end{align}

### Acknowledgment

This problem was kindly posted at the CutTheKnotMath facebook page by Leo Giugiuc, along with two solutions (Solutions 1 and 2 above) of his. The problem has been offered to Grade 9 students at the 2018 Romanian Mathematical Olympiad.

Solution 3 is by Sourabh Dubey.