Inequality Polynomial Product
For a positive integer $ m, $ let $ f(m) = m^2 + m + 1 $. Find the largest positive integer $ n $ such that
\[1000 f(1^2) f(2^2) \dotsm f(n^2) \ge f(1)^2 f(2)^2 \dotsm f(n)^2.\]
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- $\frac{a}{b}$
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- $a^n$
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- $\sqrt{}$
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- $\infty$
Solution
Note that $ f(k^2) = k^4 + k^2 + 1 $. By a little give and take,
\begin{align*}
f(k^2) &= (k^4 + 2k^2 + 1) - k^2 \\
&= (k^2 + 1)^2 - k^2 \\
&= (k^2 + k + 1)(k^2 - k + 1) \\
&= f(k) (k^2 - k + 1).\end{align*}Furthermore,
\[f(k - 1) = (k - 1)^2 + (k - 1) + 1 = k^2 - 2k + 1 + k - 1 = k^2 - k + 1,\]so
\[f(k^2) = f(k) f(k - 1).\]Thus, the given inequality becomes
\[1000 f(1) f(0) \cdot f(2) f(1) \cdot f(3) f(2) \dotsm f(n - 1) f(n - 2) \cdot f(n) f(n - 1) \ge f(1)^2 f(2)^2 \dotsm f(n)^2,\]which simplifies to
\[1000 \ge f(n).\]The function $ f(n) $ is increasing, and $ f(31) = 993 $ and $ f(32) = 1057, $ so the largest such $ n $ is $ \boxed{31} $.
Freeflows