# A Cyclic Inequality from the 6th IMO, 1964

### Solution 1

By substituting $a=x+y,$ $b=y+z,$ $c=z+x,$ $(x,y,z\gt 0)$ the given inequality becomes

$6xyz \le x^2y+xy^2+y^2z+yz^2+z^2x+zx^2$

which follows immediately by the AM-GM inequality.

### Solution 2

\displaystyle \begin{align} &\sum_{cycl}a^2(b+c-a)\le 3abc &\Leftrightarrow\\ &\sum_{cycl}[a^3-a^2(b+c)+abc]\ge 0 &\Leftrightarrow\\ &\sum_{cycl}a(a^2-a(b+c)+bc)\ge 0 &\Leftrightarrow\\ &\sum_{cycl}a(a-b)(a-c)\ge 0. \end{align}

Due to symmetry, we can assume, WLOG, $a\ge b\ge c\ge 0.$ Then

$a(a-b)(a-c)\ge 0$ and $c(c-a)(c-b)\ge 0.$

Further

\displaystyle \begin{align} \sum_{cycl}a(a-b)(a-c)&=a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\\ &\ge b(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\\ &= b(a-b)[(a-c)-(b-c)]+c(c-a)(c-b)\\ &= b(a-b)^2+c(c-a)(c-b)\ge 0. \end{align}

### CD3 Theorem

Let $P(a,b,c)$ be a cyclic homogeneous polynomial of degree $3.$ The inequality $P\ge 0$ holds for all non-negative $a,b,c$ if and only if

$P(1,1,1)\ge 0$ and
$P(a,b,0)\ge 0,$ for all $a,b\ge 0.$

Let's define $\displaystyle P(a,b,c)=3abc-\sum_{cycl}a^2(b+c-a).$ Then $P(1,1,1)=3-3=0.$ In addition,

\begin{align}P(a,b,0)&=-[a^2(b-a)+b^2(a-b)]=(a^2-b^2)(a-b)\\ &=(a-b)^2(a+b)\ge 0. \end{align}

As we see, the inequality holds for all $a,b,c\ge 0,$ not just the side lengths of a triangle, making that stipulation a red herring.

The inequality $P(a,b,0)\ge 0$ becomes equality for $a=b.$ We may want to investigate the original inequality under this condition:

\displaystyle\begin{align} P(a,a,c) &=3a^2c-[a^2c+a^2c+c^2(2a-c)]\\ &=a^2c-2ac^2+c^3=c(a^2-2ac+c^2)=c(a-c)^2, \end{align}

which attains its minimum of $0$ for $a=c$ and $c=0.$ Thus in the problem, equality is attained for $(a,a,a)$ and $(a,a,0)$ and permutations.

### Solution 4

The required inequality would nowadays be simply referred to as Schur's and accepted without a proof.

### Acknowledgment

This is problem #2 from the Sixth IMO, 1964. Solution 1 is from The IMO Compendium by D. Djukić et al; Solution 2 is by Marian Dinca.