Very easy problem! Schur's inequality $x^3+y^3+z^3+3xyz\geq x^2y+xy^2+y^2z+yz^2+z^2x+zx^2$ for $x=ab^2, \ y=bc^2, \ z=ca^2$ gives $\begin{aligned} a^3b^6+b^3c^6+c^3a^6 + 3a^3b^3c^3 &\geq a^2b^5c^2+ab^4c^4+b^2c^5a^2+bc^4a^4+c^2a^5b^2+ca^4b^4 \\ &= \left(a^2b^5c^2+b^2c^5a^2+c^2a^5b^2\right)+\left(ab^4c^4+bc^4a^4+ca^4b^4\right) \\ &= a^2b^2c^2\left(a^3+b^3+c^3\right)+abc\left(a^3b^3+b^3c^3+c^3a^3\right) \end{aligned}$ which is the above inequality!

yea!! really too trivial for an international competition! and sorry for posting same solution ! schur is insta-solve! my solution= by schur of degree $1$ we have $u^3+v^3+w^3+3uvw\ge \sum uv^2 +\sum u^2v$ so setting $u=ab^2,v=bc^2,w=ca^2$ gives $a^3b^6+b^3c^6+c^3a^6+3a^3b^3c^3\ge a^2b^5c^2+a^2b^2c^5+a^5b^2c^2+ab^4c^4+a^4bc^4+a^4b^4c$ which is equivalent to \begin{align*} a ^ 3b ^ 6 + b ^ 3c ^ 6 + c ^ 3a ^ 6 + 3a ^ 3b ^ 3c ^ 3 &\ge{ abc \left (a ^ 3b ^ 3 + b ^ 3c ^ 3 + c ^ 3a ^ 3 \right) + a ^ 2b ^ 2c ^ 2 \left (a ^ 3 + b ^ 3 + c ^ 3 \right)}. \end{align*} so we are done

$INEQ\Leftrightarrow \frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}+3\geq \frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}$ Let $\frac{a}{b}=x,\frac{b}{c}=y,\frac{c}{a}=z$,so $xyz=1$.And we have to prove that: $x^3+y^3+z^3+3\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\Leftrightarrow x^3+y^3+z^3+3xyz\geq \sum xy(x+y)$ (because $xyz=1$) So we are done!

It is just substituting and then Schur Inequality...

Yeah @Wangzu, My solution the same as your's one.

Setting in $Schurs$ Inequality $x=a^2b$ and cyclic you get following inequality. Very easy for BMO. Montenegro doesn't know to give good problem.

Montenegro is without actual blame; they just proposed a (bad) problem. The onus was on the International Jury not to have selected it - nobody twisted their hand.

RIP, mavropnevma. I salute you for your contributions to AoPS. Of course, not seeing the quick solution does not make it too much harder. If $abc = 0$, it is clear. So else we can normalize $abc = 1$. We find it equivalent to $\sum_{cyc}\frac{a^3}{b^3} + 3 \ge \sum_{cyc}\left(a^3 + \frac{1}{a^3}\right)$. We will show that $\sum_{cyc}\frac{a}{b} + 3 \ge \sum_{cyc}\left(a + \frac{1}{a}\right)$ if $abc = 1$, and this will clearly suffice. Now either two of these are $\ge 1$ and one is $\le 1$ or two of these are $\le 1$ and one is $\ge 1$. Since the inequality is cyclic, we can assume that $a, b$ are on the same side of $1$. Now using $c = \frac{1}{ab}$, this is equivalent to $\frac{a}{b} + ab^2 + \frac{1}{a^2b} + 3 \ge a + \frac{1}{a} + b + \frac{1}{b} + ab + \frac{1}{ab}$, or just the polynomial inequality $a^3b^3 + a^3 + 3a^2b + 1 \ge a^3b^2 + a^3b + a^2b^2 + a^2 + ab + a$. Lemma. If $x, y \ge 1$ then $x^3y^3 + x^3 + 3x^2y + 1 \ge x^3y^2 + x^3y + x^2y^2 + x^2 + xy + x$. Proof: Indeed, simply write $x = 1 + d, y = 1 + e$ to find it equivalent to a polynomial with all nonnegative coefficients in $d, e$ being nonnegative. We know that either $a, b \ge 1$ or $a, b \le 1$. So in the first case apply the Lemma for $x = a, y = b$; in the second case, apply it for $x = \frac{1}{b}, y = \frac{1}{a}$.

For storage purposes: Well, this is odd, it's just schur's 3rd degree inequality for $(ab^2,bc^2,ca^2)$, \begin{align*}(ab^2)^3+(bc^2)^3+(ca^2)^3+3(ab^2)(bc^2)(ca^2)&=a ^ 3b ^ 6 + b ^ 3c ^ 6 + c ^ 3a ^ 6 + 3a ^ 3b ^ 3c ^ 3 \\ &\ge a^5b^2c^2+a^2b^5+a^2b^2c^5+ab^4c^4+a^4bc^4+a^4b^4c \\ &\ge{ abc \left (a ^ 3b ^ 3 + b ^ 3c ^ 3 + c ^ 3a ^ 3 \right) + a ^ 2b ^ 2c ^ 2 \left (a ^ 3 + b ^ 3 + c ^ 3 \right)}. \quad \blacksquare \end{align*}

Let $x=ab^2,y=bc^2,z=ca^2$, then we want to show that $x^3+y^3+z^3+3xy\ge\sum_{\text{sym}}xy^2$ which is Schur.

Does this solution work, Note that the sequence (3,3,3) majorizes (2,2,5), from muirhead using this fact we have $3a^3b^3c^3\ge a^2b^2c^2(a^3+b^3+c^3)$, Also note that the sequence (6,3,0) majorizes (4,4,1) again by muirhead we know $a ^ 3b ^ 6 + b ^ 3c ^ 6 + c ^ 3a ^ 6\ge (abc)(a ^ 3b ^ 3 + b ^ 3c ^ 3 + c ^ 3a ^ 3)$, Can someone check if this solution works

mathlete5451006 wrote: Does this solution work, Note that the sequence (3,3,3) majorizes (2,2,5), from muirhead using this fact we have $3a^3b^3c^3\ge a^2b^2c^2(a^3+b^3+c^3)$, Also note that the sequence (6,3,0) majorizes (4,4,1) again by muirhead we know $a ^ 3b ^ 6 + b ^ 3c ^ 6 + c ^ 3a ^ 6\ge (abc)(a ^ 3b ^ 3 + b ^ 3c ^ 3 + c ^ 3a ^ 3)$, Can someone check if this solution works (3,3,3) doesn't majorize (5,2,2).

wait so you cannot just rearrange the terms from the sequence to make one majorize the other?

$(3,3,3)\not\succ(2,2,5)$ since $3<5$.

Shur inequality

mathlete5451006 wrote: wait so you cannot just rearrange the terms from the sequence to make one majorize the other? The sequence should be descending order.

Set $x=ab^2$ etc. we wish to prove $$x^3+y^3+z^3+3xy\geqslant x^2y+xy^2+x^2z+xz^2+y^2z+yz^2$$this is Schur.

A sure nuke!

Coll. Math. J., 20 No.4 (Sept., 1989) 343. 406. Proposed by M. S. Klamkin, University of Alberta, Edmonton Prove that$$\frac{a^3}{b^3}+\frac{b^3}{a^3}+\frac{b^3}{c^3}+\frac{c^3}{b^3}+\frac{c^3}{a^3}+\frac{a^3}{c^3}\ge \frac{a^2}{bc}+\frac{bc}{a^2}+\frac{b^2}{ca}+\frac{ca}{b^2}+\frac{c^2}{ab}+\frac{ab}{c^2}$$where $a,b,c> 0.$

A bit longer for #1 but: Let $a^3 = \frac{x}{y}, b^3 = \frac{y}{z}, c^3 = \frac{z}{x}$. Then we get that the inequality is equivalent to: $\frac{xy}{z^2} + \frac{xz}{y^2} + \frac{yz}{x^2} + 3 > \frac{x}{z} + \frac{y}{x} + \frac{z}{y} + \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \iff \sum_{\text{sym}} x^3y^3 + 3x^2y^2z^2 > \sum_{\text{sym}} x^3y^2z.$ Now let $xy = m, yz = n, zx = p$. Then we get $\sum_{\text{sym}} m^3 + 3mnp > \sum_{\text{sym}} m^2n$, which is true by Schur.

Let $x=a^2b, y=b^2c, z=c^2a$ Now the inequality is $$x^3+y^3+z^3+3xyz\geq x^2y+y^2z+z^2x+xy^2+yz^2+zx^2$$and it is true from Schur's inequality ($(3,0,0)+(1,1,1)\geq 2\cdot(2,1,0)$)

Replacing $x,y,z\in \mathbf{R^+}$ with $x,y,z\geq 0$ nonnegative reels, the inequality is still true.

Let's donate $ab^2=x$ $bc^2=y$ $a^2c=z$ $x^3+y^3+z^3+3xyz \geq x^2y+y^2z+z^2x+x^2z+z^2y+y^2x$ which is schur inequality

ismayilzadei1387 wrote: Let's donate $ab^2=x$ $bc^2=y$ $a^2c=z$ $x^3+y^3+z^3+3xyz \geq x^2y+y^2z+z^2x+x^2z+z^2y+y^2x$ which is schur inequality yes! this it is post #2 !