IMO 2012 #2

by Wolstenholme, Oct 28, 2014, 6:01 PM

Let $n\ge 3$ be an integer, and let $a_2,a_3,\ldots ,a_n$ be positive real numbers such that $a_{2}a_{3}\cdots a_{n}=1$. Prove that
\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]

Proof:

Note that for all integer $ 2 \le i \le n $ we have by AM - GM that $ (1 + a_i)^i = \left(\frac{1}{i - 1} + \frac{1}{i - 1} + \dots + \frac{1}{i - 1} + a_i\right)^i \ge \frac{i^ia_i}{(i - 1)^{i - 1}}. $ Therefore $ \prod_{i = 2}^{n}(1 + a_i)^i \ge \prod_{i = 2}^{n}\frac{i^ia_i}{(i - 1)^{i - 1}} = n^na_2a_3a_4\dots{a_n} = n^n $. Equality holds only if $ a_i = \frac{1}{i - 1} $ for all integer $ 2 \le i \le n $ but as this is clearly impossible, we are done.

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  • HW$\textcolor{white}{}$

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  • Uh-oh ARML practice is Thursday... I should start the homework. :P

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  • Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial :P

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  • Hi Wolstenholme did you actually use calc on that tstst problem :o

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