Olympiad Maths Prep

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Problem 1663

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.7 Prove it

□\square Example 8 Given the sequence {an}\left\{a_{n}\right\} satisfies ∘a1=2116,2an−3an−1=32n+1,n⩾2{ }^{\circ} a_{1}=\frac{21}{16}, 2 a_{n}-3 a_{n-1}=\frac{3}{2^{n+1}}, n \geqslant 2, let mm be a positive integer, m⩾2m \geqslant 2. Prove: when n⩽mn \leqslant m, we have (an+32n+3)1m\left(a_{n}+\frac{3}{2^{n+3}}\right)^{\frac{1}{m}}. [m−(23)m(m−1)m]<m2−1m−n+1\left[m-\left(\frac{2}{3}\right)^{\frac{m(m-1)}{m}}\right]<\frac{m^{2}-1}{m-n+1}. (2005 China Mathematical Olympiad Problem)

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

From the given conditions, we have 2nan=3⋅2n−1an−1+342^n a_n = 3 \cdot 2^{n-1} a_{n-1} + \frac{3}{4}. Let bn=2nanb_n = 2^n a_n, for n=1,2,3,⋯n = 1, 2, 3, \cdots, then bn=3bn−1+34b_n = 3 b_{n-1} + \frac{3}{4}, which means
bn+38=3(bn−1+38) b_n + \frac{3}{8} = 3 \left( b_{n-1} + \frac{3}{8} \right)

Since b1=2a1=218b_1 = 2 a_1 = \frac{21}{8}, we have bn+38=3n−1(b1+38)=3nb_n + \frac{3}{8} = 3^{n-1} \left( b_1 + \frac{3}{8} \right) = 3^n, thus
an=(32)n−32n+3 a_n = \left( \frac{3}{2} \right)^n - \frac{3}{2^{n+3}}

Therefore, to prove the original inequality, it suffices to prove
(32)nm⋅[m−(23)n(m−1)m]<m2−1m−n+1 \left( \frac{3}{2} \right)^{\frac{n}{m}} \cdot \left[ m - \left( \frac{2}{3} \right)^{\frac{n(m-1)}{m}} \right] < \frac{m^2 - 1}{m - n + 1}

That is, it suffices to prove
(1−nm+1)(32)nm⋅[m−(23)n(m−1)m]<m−1 \left( 1 - \frac{n}{m+1} \right) \left( \frac{3}{2} \right)^{\frac{n}{m}} \cdot \left[ m - \left( \frac{2}{3} \right)^{\frac{n(m-1)}{m}} \right] < m - 1

First, estimate the upper bound of 1−nm+11 - \frac{n}{m+1}. By Bernoulli's inequality, we have 1−nm+1<(1−1m+1)n1 - \frac{n}{m+1} < \left( 1 - \frac{1}{m+1} \right)^n, so
(1−nm+1)m<(1−1m+1)mn=(mm+1)mn=[1(1+1m)m]n \left( 1 - \frac{n}{m+1} \right)^m < \left( 1 - \frac{1}{m+1} \right)^{mn} = \left( \frac{m}{m+1} \right)^{mn} = \left[ \frac{1}{\left( 1 + \frac{1}{m} \right)^m} \right]^n
(Note: Alternatively, using the AM-GM inequality, the number of 1's is mn−mmn - m, then
(1−nm+1)m=(1−nm+1)m⋅1⋅1⋅⋯⋅1<[m(1−nm+1)+mn−mmn]mn=(mm+1)mn \begin{aligned} & \left( 1 - \frac{n}{m+1} \right)^m = \left( 1 - \frac{n}{m+1} \right)^m \cdot 1 \cdot 1 \cdot \cdots \cdot 1 \\ < & \left[ \frac{m \left( 1 - \frac{n}{m+1} \right) + mn - m}{mn} \right]^{mn} = \left( \frac{m}{m+1} \right)^{mn} \end{aligned}

Since m≥2m \geq 2, by the binomial theorem, we have
(1+1m)m≥1+Cm1⋅1m+Cm2⋅1m2=52−12m≥94 \left( 1 + \frac{1}{m} \right)^m \geq 1 + C_m^1 \cdot \frac{1}{m} + C_m^2 \cdot \frac{1}{m^2} = \frac{5}{2} - \frac{1}{2m} \geq \frac{9}{4}

Thus, (1−nm+1)m<(49)n\left( 1 - \frac{n}{m+1} \right)^m < \left( \frac{4}{9} \right)^n, i.e., 1−nm+1<(23)2nm1 - \frac{n}{m+1} < \left( \frac{2}{3} \right)^{\frac{2n}{m}}. Therefore, to prove (1), it suffices to prove

i.e.,
(23)2nm(32)nm⋅[m−(23)n(m−1)m]<m−1 \left( \frac{2}{3} \right)^{\frac{2n}{m}} \left( \frac{3}{2} \right)^{\frac{n}{m}} \cdot \left[ m - \left( \frac{2}{3} \right)^{\frac{n(m-1)}{m}} \right] < m - 1
(23)nm⋅[m−(23)n(m−1)m]<m−1 \left( \frac{2}{3} \right)^{\frac{n}{m}} \cdot \left[ m - \left( \frac{2}{3} \right)^{\frac{n(m-1)}{m}} \right] < m - 1

Let (23)nm=t\left( \frac{2}{3} \right)^{\frac{n}{m}} = t, then 0<t<10 < t < 1, (2) becomes t(m−tm−1)<m−1t \left( m - t^{m-1} \right) < m - 1, i.e.,
(t−1)[m−(tm−1+tm−2+⋯+t+1)]<0 (t - 1) \left[ m - \left( t^{m-1} + t^{m-2} + \cdots + t + 1 \right) \right] < 0

This inequality is clearly true, thus the original inequality holds.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.