# An Inequality from the 1967 IMO Shortlist

### Solution 1

The inequality is equivalent to

$a^2b^3c^3+a^3b^2c^3+a^3b^3c^2\le a^8+b^8+c^8.$

Since the sequence $(8,0,0)\,$ majorizes the sequence $(2,3,3),\,$ this is a direct consequence of Muirhead's inequality.

### Solution 2

By the AM-GM inequality,

\begin{align} 8a^2b^3c^3&\le 2a^8 + 3b^8+3c^8\\ 8a^3b^2c^3&\le 3a^8 + 2b^8+3c^8\\ 8a^3b^3c^2&\le 3a^8 + 3b^8+2c^8 \end{align}

Summing up and dividing by $8a^3b^3c^3\,$ solves the problem.

### Solution 3

\displaystyle \begin{align} &f=\frac{a^8+b^8+c^8}{a^3 b^3 c^3}-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 0\\ &\frac{a^8+b^8+c^8-a^3 b^3 c^2-a^3 b^2 c^3-a^2 b^3 c^3}{a^3 b^3 c^3} \geq\\ &f_1=a^8+b^8+c^8-a^3 b^3 c^2-a^3 b^2 c^3-a^2 b^3 c^3 \geq 0 \end{align}

The problem is that there appears to be no standard proof for the arrangement inequality for three sequences, however, there is a reference. But no problemo:

$a \geq b \geq c$ wlog, so let us set $b= c+\epsilon ,a=c+\delta +\epsilon$, $\epsilon, \delta \geq 0$.

$f_1=c^8-c^3 (c+\epsilon )^2 (c+\delta +\epsilon )^3-c^3 (c+\epsilon )^3 (c+\delta +\epsilon )^2-\\ \qquad\qquad\qquad c^2 (c+\epsilon )^3 (c+\delta +\epsilon )^3+(c+\delta +\epsilon )^8+(c+\epsilon )^8$

By expanding we miraculously get the negative terms out of the way and can resume daily life:

$21 c^6 \left(\delta ^2+\delta \epsilon +\epsilon ^2\right)+18 c^5 (\delta +2 \epsilon ) \left(3 \delta ^2+2 \delta \epsilon +2 \epsilon ^2\right)\\ \qquad\qquad\qquad +5 c^4 \left(14 \delta ^4+55 \delta ^3 \epsilon +78 \delta ^2 \epsilon ^2+46 \delta \epsilon ^3+23 \epsilon ^4\right)\\ \qquad\qquad\qquad +4 c^3 (\delta +2 \epsilon ) \left(14 \delta ^4+42 \delta ^3 \epsilon +55 \delta ^2 \epsilon ^2+26 \delta \epsilon ^3 +13 \epsilon ^4\right)\\ \qquad\qquad\qquad+c^2 \left(28 \delta ^6+168 \delta ^5 \epsilon +420 \delta ^4 \epsilon ^2+559 \delta ^3 \epsilon ^3+417 \delta ^2 \epsilon ^4+165 \delta \epsilon ^5+55 \epsilon ^6\right)\\ \qquad\qquad\qquad +8 c (\delta +2 \epsilon ) \left(\delta ^6+5 \delta ^5 \epsilon+11 \delta ^4 \epsilon ^2+13 \delta ^3 \epsilon ^3+9 \delta ^2 \epsilon ^4+3 \delta \epsilon ^5+\epsilon ^6\right)\\ \qquad\qquad\qquad +\delta ^8+8 \delta ^7 \epsilon+28 \delta ^6 \epsilon ^2+56 \delta ^5 \epsilon ^3+70 \delta ^4 \epsilon ^4+56 \delta ^3 \epsilon ^5+28 \delta ^2 \epsilon ^6+8 \delta \epsilon ^7+2 \epsilon ^8$

### Solution 4

$a^2b^2c^2(ab+bc+ca)\le a^8+b^8+c^8. First off, by the AM-GM inequality, \displaystyle$\displaystyle a^2b^2c^2=\sqrt[3]{a^6b^6c^6}\le\frac{a^6+b^6+c^6}{3}.$Also,$ab+bc+ca\le a^2+b^2+c^2.\,$Thus suffice it to prove that$\displaystyle \frac{(a^6+b^6+c^6)(a^2+b^2+c^2)}{3}\le \frac{3(a^8+b^8+c^8)}{3}.$This reduces to$(a^6b^2+b^6c^2+c^6a^2)+(a^2b^6+b^2c^6+c^2a^6)\le 2(a^8+b^8+c^8).$The latter follows by the Rearrangement inequality, applied twice:$\begin{align}a^6b^2+b^6c^2+c^6a^2&\le a^8+b^8+c^8\\ a^2b^6+b^2c^6+c^2a^6&\le a^8+b^8+c^8. \end{align}\$

### Acknowledgment

I found this problem in IMO Compendium (1st edition, p 46). The problem (proposed by Poland) was shortlisted at the 9th IMO in 1967. Solution 3 is by N.N. Taleb; Solution 4 is by Adrian Muresan.