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If $a^2b^2+b^2c^2+c^2a^2-69abc=2016$, then, what can be said about the least value of $\min(a, b ,c)$, where $a, b, c\gt0$?

This problem is unyielding to the major inequalities like AM-GM, Cauchy-Schwarz, etc. I also tried relating it to $x^3+y^3+z^3-3xyz=(x+y+z)(\sum_{cyc}x^2+\sum_{cyc}xy)$, but of no use. Any ideas. Thanks beforehand.

PS: This is problem S395 in Issue 6 2016 of Mathematical Reflections.

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    Can $a,b,c<0$ or just looking at $a,b,c\geq 0$ – 2017-01-10
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    @kingW3 yes, $a,b,c$ are positive – 2017-01-10
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    Is this from an ongoing contest? See comment [there](http://math.stackexchange.com/q/2092988/). – 2017-01-11
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    Deleting this. Undeleting on Jan 16th – 2017-01-12

1 Answers 1

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Let $a,c \in \mathbb{R}_{>0}$. Assume $a = \epsilon > 0$ and $c = 1$. This leads to the quadratic equation in $b$

$$(1+\epsilon^2)b^2-69\epsilon b+\epsilon^2-2016 = 0,$$

with a solution given by

$$ b = \frac{69\epsilon +\sqrt{8064+12821\epsilon^2 -4\epsilon^4}}{2(1+\epsilon^2)} $$

Thus, $b \in \mathbb{R}_{>0}$ if and only if $8064+12821\epsilon^2 -4\epsilon^4 \geq 0$, which is true for $0 < \epsilon \leq 56.6205...$.

Thus, it is easy to conclude about $\min (a,b,c)$.

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    so then is the minimum value 1? – 2017-01-10
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    Nope, you can make $\epsilon$ as small as you want $(\epsilon \rightarrow 0)$ so that the conclusion is that there is no minimum. – 2017-01-10
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    Which means $\text{inf(min(a,b,c)=0})$, isnt it? – 2017-01-10
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    Yeah! zero is the infimum. – 2017-01-10