2018 UMO Problems/Problem 2

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Problem 2

Let $P(x)$ be a cubic polynomial $(x-a)(x-b)(x-c)$, where $a, b,$ and $c$ are positive real numbers. Let Q(x) be the polynomial with $Q(x) = (x-ab)(x-bc)(x-ac)$. If $P(x) = Q(x)$ for all $x$, then find the minimum possible value of $a + b + c$.

Solution 1

Plugging in $x = 0$, we find that $abc = 1$. Using AM-GM, we have that $a+b+c \leq 3 \sqrt[3]{abc} = \fbox{3}$ ~bigbrain123