Eukleidis wrote: Problem 1. Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that: $\displaystyle\prod(a^5+a^4+a^3+a^2+a+1)\geq 8(a^2+a+1)(b^2+b+1)(c^2+c+1)$ It is Holder with $(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3=8$ (first a typo)

Note that $a^5 +a^4 +a^3 +a^2 +a +1=\frac{a^6 -1}{a-1}$, and $(a-1)(a^2 +a+1)=a^3 -1$. Also, $a^6 -1=(a^3 -1)(a^3 +1)$, so we actually have to prove that: \[ (a^3 +1)(b^3 +1)(c^3 +1) \geqslant 8 \] By AM-GM we have that: \[ \frac{1}{a^3 +1} +\frac{1}{b^3 +1} +\frac{1}{c^3 +1} \geqslant \frac{3}{\sqrt[3]{(a^3 +1)(b^3 +1)(c^3 +1)}} \] and \[ \frac{a^3}{a^3 +1} +\frac{b^3 }{b^3 +1} +\frac{c^3 }{c^3 +1} \geqslant \frac{3abc}{\sqrt[3]{(a^3 +1)(b^3 +1)(c^3 +1)}} \] Adding the last two inequalities side by side we get: \[ 3 \geqslant \frac{3+3abc}{\sqrt[3]{(a^3 +1)(b^3 +1)(c^3 +1)}} \] Taking into account that $abc=1$, we have: \[{ \sqrt[3]{(a^3 +1)(b^3 +1)(c^3 +1)}} \geqslant 2 \] \[ (a^3 +1)(b^3 +1)(c^3 +1) \geqslant 8 \] $\mathbb{Q}. \mathbb{E}. \mathbb{D}$.

We see the inequality is true for $a=b=c=1$ because $6^3 \geq8*3^3$. Noting that $a^n+a^{n-1}+...+a+1=\frac{a^{n+1}-1}{a-1}$ and that $\frac{a^6-1}{a^3-1}=a^3+1$ the problem simplifies to proving that: $(a^3+1)(b^3+1)(c^3+1) \geq 8$ or that $a^3+\frac{1}{a^3}+b^3+\frac{1}{b^3}+c^3+\frac{1}{c^3} \geq 6$ which is directly true by AM-GM.

Since $x^5+x^4+x^3+x^2+x+1 = (x^3+1)(x^2+x+1)$ inequality is eqvivalent to this one $(a^3+1)(b^3+1)(b^3+1)\geq 8$ witch is obviusly true since $x^3+1\geq 2\sqrt{x^3}$ is true by AM-GM.

SCP wrote: It is Holder with $(a^3+1)(b^3+1)(c^3+1)\ge (\sqrt[3]{abc}+1)^3=8$ By the way, Hölder is $(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3=8$. It is incredible how a problem, only because it was (wrongly, I think - it is so obvious in all steps, that it trivializes the competition) sat in a contest, induced such complicated solutions as most in the above. The factorization $x^5 + x^4 + x^3 + x^2 + x + 1 = (x^3+1)(x^2+x+1)$ is forced, and $x^3+1 \geq 2\sqrt{x^3}$ so direct, that it's almost embarassing. Sorry for repeating almost verbatim the post just above (a benign typo has no real relevance), but I could not resist emphasizing its immediate and natural flavour.

An epiphany. Why three variables $a,b,c$ ? With $n$ variables, the proof is identical, with the constant being $2^n$; aha, it will give the contestants an idea on what to do, right? so let's ask it for $n=3$, and write $8$ instead of $2^3$. If forced (with a gun at the temple of my head) to cast a yea vote for this problem, the more honest presentation would have been for $n=1$ (why not a one variable inequality if we are prepared to sink to these depths) Show that for a real number $a \geq 1$ the inequality $\displaystyle a^5+a^4+a^3+a^2+a+1 \geq 2(a^2+a+1)$ holds.

mavropnevma wrote: Show that for a real number $a \geq 1$ the inequality $\displaystyle a^5+a^4+a^3+a^2+a+1 \geq 2(a^2+a+1)$ holds. Maybe because they only would say $a^5\ge a^2$ and so on.

$(a^3+1)(b^3+1)(c^3+1)>=(2a\sqrt{a})(2b\sqrt{b})(2c\sqrt{c})=8.$ This inequality is very weak.

Alternatively, just use Muirhead. It's just $(4,4,1) + (5,2,2)\geq\ 2(3,3,3)$ which is trivial.

Eukleidis wrote: Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that: $\displaystyle\prod(a^5+a^4+a^3+a^2+a+1)\geq 8(a^2+a+1)(b^2+b+1)(c^2+c+1)$ See(it's generalization): http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=420083

actually,we only have to notice $a^5+a^4+a^3+a^2+a+1\ge 2(a^{\frac{7}{2}}+a^{\frac{5}{2}}+a^{\frac{3}{2}})$.

note that , $a^5+a^4+a^3+a^2+a+1=a^3(a^2+a+1)+(a^2+a+1)$ $=(a^3+1)(a^2+a+1)\ge 2a^{3/2}.(a^2+a+1)$ so , $\prod[a^5+a^4+a^3+a^2+a+1]\ge8\prod(a^2+a+1)$,as $abc=1$

Number1 wrote: Since $x^5+x^4+x^3+x^2+x+1 = (x^3+1)(x^2+x+1)$ inequality is eqvivalent to this one $(a^3+1)(b^3+1)(b^3+1)\geq 8$ witch is obviusly true since $x^3+1\geq 2\sqrt{x^3}$ is true by AM-GM. Nice

Eukleidis wrote: Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that: $\displaystyle\prod(a^5+a^4+a^3+a^2+a+1)\geq 8(a^2+a+1)(b^2+b+1)(c^2+c+1)$ See also here

Let $a,b,c$ be positive real numbers such that $abc\geq 1$. Prove that$$8(a^2-a+1)(b^2-b+1)(c^2-c+1)\geq (a+1)(b+1)(c+1)$$

$a^5+a^4+a^3+a^2+a+1=(a^3+1)(a^2+a+1)$, so it remains to prove that $(a^3+1)(b^3+1)(c^3+1)\geq 8$, or $(a^3+1)^{1/3}(b^3+1)^{1/3}(c^3+1)^{1/3}\geq 2$. Doesn't Holder's inequality apply by $LHS\geq (a^3b^3c^3)^{1/3}+(1*1*1)^{1/3}=2$?

\begin{align*} \prod(a^5+a^4+a^3+a^2+a+1) &\geq 8\prod(a^2+a+1)\iff\\ \iff \prod(a^3+1)(a^2+a+1) &\geq 8\prod(a^2+a+1)\iff\\ \iff \prod(a^3+1) &\geq 8.\\ \end{align*}$$\prod(a^3+1) = a^3b^3c^3+a^3b^3+b^3c^3+c^3a^3+a^3+b^3+c^3+1 \stackrel{\mathrm{AM-GM}}{\geq} 8\sqrt[8]{(abc)^{12}\cdot1}=8. \quad\square $$