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Algebra Difficulty 7.4 National olympiad, round 2 Prove it

Example 8 Given the sequence {an}\left\{a_{n}\right\} satisfies a1=2116a_{1}=\frac{21}{16},
2an−3an−1=32n+1,n⩾22 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[m−(23)n(m−1)m]<m2−1m−n+1\left(a_{n}+\frac{3}{2^{n+3}}\right)^{\frac{1}{m}}\left[m-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right]<\frac{m^{2}-1}{m-n+1}

Solution

Prove that from equation (1) we get 2nan=3⋅2n−1an−1+342^{n} a_{n}=3 \cdot 2^{n-1} a_{n-1}+\frac{3}{4}, let bn=2nan,n=1,2,3,⋯b_{n}=2^{n} a_{n}, n=1,2,3, \cdots
bn=3bn−1+34bn+38=3(bn−1+38)\begin{array}{c} b_{n}=3 b_{n-1}+\frac{3}{4} \\ b_{n}+\frac{3}{8}=3\left(b_{n-1}+\frac{3}{8}\right) \end{array}

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}, hence
an=(32)n−32n+3a_{n}=\left(\frac{3}{2}\right)^{n}-\frac{3}{2^{n+3}}

Therefore, to prove equation (2), 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}

It suffices to prove □\square
(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}

Thus,
(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)^{m n}=\left(\frac{m}{m+1}\right)^{m n}=\left[\frac{1}{\left(1+\frac{1}{m}\right)^{m}}\right]^{n}
(Note: This can also be derived using the AM-GM inequality: 1 appears mn−mm n-m times (1−nm+1)m=\left(1-\frac{n}{m+1}\right)^{m}=
(1−nm+1)m⋅1⋅1⋅⋯⋅1<[m(1−nm+1)+mn−mmn]mn=(mm+1)mn.)\left.\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)+m n-m}{m n}\right]^{m n}=\left(\frac{m}{m+1}\right)^{m n} .\right)

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

Therefore, (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{2 n}{m}}

Thus, to prove equation (3), it suffices to prove
(23)2nm(32)nm⋅(m−(23)n(m−1)m)<m−1\left(\frac{2}{3}\right)^{\frac{2 n}{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

i.e.,
(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, equation (4) 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, hence the original inequality holds.

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