Would you have to prove that $x^3\ge y^3\Rightarrow x\ge y$? I did this just to be on the safe side...

No; the last step uses the fact that if $xyz \le w^3$ then one of $x,y,z$ must be less than or equal to $w$; this is clear because if $x,y,z > w$, $xyz > w^3$, contradiction.

No since it's obvious enough to the graders I think.

Oh well I said $\text{min}^3\le\text{product of the terms}\le (abc)^3\Rightarrow \text{min}\le abc$ because of the abovestated fact

What a stupid inequality.

AkshajK wrote:

What's the motivation behind "noticing" that?

Basically, I wanted to do contradiction and the cyclicness made it ugly so I assumed that everything >abc and then multiplied all of them together so it wasn't ugly anymore, isolated variables and done.

Well the first thing I thought was wow the LHS is really annoying in that each term only has two of the variables and there's also the minimum condition. Then I looked for equality case so I set a=b=c which ended up giving me $(a-1)^5$. This kind of motivates AkshajK's solution. Also, as I said the minimum condition is really annoying, and you can't really add the LHS so you might as well multiply.

Notice 1,5,10,5,1 in some order. Does this look familiar? Is this a coincidence?

Yeah finding the equality case was what gave me the idea. Also I originally substituted $x=a-1$, $y=b-1$, $z=c-1$ which made it $x^5 \ge 0$, a lot easier to see.

Darn the $a, b, c \ge 1$ was a total giveaway that I did not catch

Hmm so my first thought was "sum all" for some reason and I rejected product out of hand at first... In all honesty I think this is a horrible problem for any olympiad.

I'm also pretty mad at this problem. The first thing I checked was equality case and I noticed the coefficients of $(x-1)^5$ and I thought sum and wasted around 2 hours before trying product.

hmm it seems I'm the only one who likes a new kind of inequality for a change...

You're not the only one. This was my favorite problem.

Wait no... any inequality that's just a special case of a generalization where the generalization was easier should not appear. I mean, what if the problem had been this: Hypothetical JMO1 wrote: Let $x_1,x_2,\ldots , x_n\ge 1$ be reals. Prove that \[\text{min} \left(\frac{10x_i^2-5x_i+1}{x_{i+1}^2-5x_{i+1}+10}\right) \le x_1^{\frac{3}{n}}x_2^{\frac{3}{n}}\ldots x_n^{\frac{3}{n}},\] where $x_{n+1}=x_1$. Now the problem is even more immediate; it's obvious to consider each separately. It's like altering a bound to be weaker than necessary to hide the solution path in a combo problem. EDIT: Fixed.

wait your inequality has a weaker bound too because the RHS should be $(x_1x_2...x_n)^{\frac{3}{n}}$ but point taken but IMO this is better than having some am-GM or Cauchy thing again

Pretty much the same, but just for storage

Probably one of the nicest JMO inequalities I've ever seen. Claim: $\frac{10a^2-5a+1}{a^2-5a+10} \leq a^3$ Proof: This rearranges into $(a-1)^5 \geq 0$, which is clearly true. Similar inequalities also hold true for $b$ and $c$. Now, just multiply all three inequalities together to get \[\left(\frac{10a^2-5a+1}{b^2-5b+10} \cdot \frac{10b^2-5b+1}{c^2-5c+10} \cdot \frac{10c^2-5c+1}{a^2-5a+10}\right )\leq (abc)^3. \] Now, the desired result is immediate. Remark: The main motivation for this solution is that minimums are really hard to work with, and WLOG to get rid of the minimum obviously won't work. Therefore, we multiply all three inequalities together, and watch the magic happen

We open with a claim. Claim 1: $\frac{10a^2-5a+1}{a^2-5a+10}\leq a^3$ Proof of Claim 1: $(a-1)^5 \geq 0 \implies a^5 - 5a^4 + 10a^3 - 10a^2 + 5a - 1 \geq 0 \implies a^5 - 5a^4 + 10a^3 \geq 10a^2 - 5a + 1 \implies \frac{10a^2-5a+1}{a^2-5a+10}\leq a^3$ Observe that $$\min \left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right )\leq \sqrt[3]{\frac{10a^2-5a+1}{b^2-5b+10}\cdot\frac{10b^2-5b+1}{c^2-5c+10}\cdot\frac{10c^2-5c+1}{a^2-5a+10}}$$$$\leq \sqrt[3]{a^3b^3c^3}$$$$\leq abc. \blacksquare$$ Remarks: My first JMO problem !

It suffices to show that: $$\frac{10a^2-5a+1}{b^2-5b+10}\cdot\frac{10b^2-5b+1}{c^2-5c+10}\cdot\frac{10c^2-5c+1}{a^2-5a+10}\le a^3b^3c^3,$$for which it suffices to show that: $$\frac{10a^2-5a+1}{a^2-5a+10}\le a^3.$$This is true since: $$a^3-\frac{10a^2-5a+1}{a^2-5a+10}=\frac{(a-1)^5}{\left(a-\frac52\right)^2+\frac{15}4}\ge0.$$

Notice \begin{align*}(a-1)^5\ge 0&\implies a^5-5a^4+10a^3-10a^2+5a-1\ge 0\\&\implies a^5-5a^4+10a^3\ge 10^2-5a+1\\&\implies a^3\ge\frac{10a^2-5a+1}{a^2-5a+10}.\end{align*}Multiplying similar inequalities with $b,c$ yields $$\prod_{\text{cyc}}\frac{10a^2-5a+1}{b^2-5b+10}=\prod_{\text{cyc}}\frac{10^2-5a+1}{a^2-5a+10}\le (abc)^3$$which implies the desired conclusion. $\square$

REMARKS The 10,-5,1 really gives away binomial theorem

First, we claim that $x^3\ge \frac{10x^2 - 5x + 1}{x^2 - 5x + 10}$ for all $x \ge 1$. Note that $x^2 - 5x + 10 > 0$, so the claim is equivalent to \[ x^5 - 5x^4 + 10x^3\ge 10x^2 - 5x + 1\iff x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1 = (x-1)^5\ge 0,\]Which is clearly true for $x\ge 1$, so the claim follows. From the claim, we have \[ \prod_{\text{cyc}}\frac{10a^2-5a + 1}{a^2 - 5a + 10}\le (abc)^3\implies \prod_{\text{cyc}}\frac{10a^2 - 5a + 1}{b^2 - 5b + 10}\le (abc)^3.\]Now, note that if $\frac{10a^2 - 5a + 1}{b^2 - 5b + 10}$ and cyclic variants were all greater than $abc$, then $\prod_{\text{cyc}}\frac{10a^2 - 5a + 1}{b^2 - 5b + 10}$ would be greater than $(abc)^3$, impossible. The desired result follows.

Claim: $a^3 \ge \frac{10a^2-5a+1}{a^2-5a+10}.$ Proof: Note, that this rearranges, to $a^3(a^2-5a+10)-10a^2+5a-1 \ge 0,$ now, $a^5-5a^4+10a^3-10a^2+5a-1=(a-1)^5 \ge 0,$ because $a \ge 1.$ Similarly, $b^3 \ge \frac{10b^2-5b+1}{b^2-5b+10}$, and $c^3 \ge \frac{10c^2-5c+1}{c^2-5c+10}.$ Now, notice, that $\frac{10a^2-5a+1}{b^2-5b+10}\cdot \frac{10b^2-5b+1}{c^2-5c+10}\cdot \frac{10c^2-5c+1}{a^2-5a+10} =\frac{10a^2-5a+1}{a^2-5a+10} \cdot \frac{10b^2-5b+1}{b^2-5b+10} \cdot \frac{10c^2-5c+1}{c^2-5c+10} \le (abc)^3$, now, note, that the minimum value of the minimum, occurs, when $\frac{10a^2-5a+1}{a^2-5a+10}=\frac{10b^2-5b+1}{b^2-5b+10} =\frac{10c^2-5c+1}{c^2-5c+10}=abc,$ and we are done.

Titu JMO algmanip p1 Claim: For $a \ge 1$, \[ (10a^2 - 5a + 1) \le a^3 \cdot (a^2 - 5a + 10) \]Proof. Follows as this is just $(a - 1)^5 \ge 0$. $\blacksquare$ Then, the result follows as \[ \prod_{\text{cyc}} \frac{10a^2-5a+1}{b^2-5b+10} = \prod_{\text{cyc}} \frac{10a^2-5a+1}{a^2-5a+10} \le (abc)^3 \]which implies the result.

I claim that for all reals $x \ge 1$, we have the following inequality: $$\frac{10x^2 - 5x + 1}{x^2 - 5x + 10} \le x^3 \iff 10x^2 - 5x + 1 \le x^5 - 5x^4 + 10x^3 \iff 0 \le \binom{5}{5}x^5 - \binom{5}{4}x^4 + \binom{5}{3}x^3 - \binom{5}{2}x^2 + \binom{5}{1}x^1 - \binom{5}{0} \iff (x - 1)^5 \ge 0,$$which is obviously true. Thus, $$\begin{aligned} \frac{10a^2 - 5a + 1}{a^2 - 5a + 10} \le a^3 \\ \frac{10b^2 - 5b + 1}{b^2 - 5b + 10} \le b^3 \\ \frac{10c^2 - 5c + 1}{c^2 - 5c + 10} \le c^3, \end{aligned}$$and multiplying all these inequalities together we get $$\left(\frac{10a^2 - 5a + 1}{a^2 - 5a + 10}\right)\left(\frac{10b^2 - 5b + 1}{b^2 - 5b + 10}\right)\left(\frac{10c^2 - 5c + 1}{c^2 - 5c + 1}\right)\le a^3b^3c^3,$$which gives the desired result. Remark. The key thing in this problem is that the first inequality is basically given to us in the problem because when putting $a = b = c$, the problem reduces to that, so in a sense our method is guaranteed to work.

We show that the product of all three is $\leq (abc)^3.$ Thus we just show $$\frac{10a^2-5a+1}{a^2-5a+10} \leq a^3,$$and after expansion this is equivalent to when $(a-1)^5 \geq 0,$ which is when $a\geq 1,$ so we're done$.\blacksquare$

Somehow came up with this instantly. Assume FTSOC that $$ \frac{10a^2-5a+1}{b^2-5b+10} < abc$$with cyclic variations holding. Multiplying all three gives us $$\prod \frac{10a^2-5a+1}{a^2-5a+10} < a^3b^3c^3$$. In fact, I claim that $$\frac{10a^2-5a+1}{a^2-5a+10} \geq a^3$$, implying the contradiction. In fact, it rearranges to $$\frac{(a-1)^5}{a^2-5a+10} \geq 0$$, done.

We will show that $$\frac{10x^2-5x+1}{x^2-5x+10}\le x^3$$for $x\ge 1$. Expanding, we want to prove that $$10x^2-5x+1\le x^5-5x^4+10x^3,$$or $$x^5-5x^4+10x^3-10x^2+5x-1=(x-1)^5\ge 0,$$which is true. Therefore, the product of the three terms in the problem is $\le a^3b^3c^3$. At least one of the terms is $\le abc$.