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Algebra Difficulty 6.6 National olympiad Prove it China

Let {an}\{a_n\} be a sequence such that a1=2116a_1 = \frac{21}{16} and
2an−3an−1=32n+1,n≥2.1◯ 2a_n - 3a_{n-1} = \frac{3}{2^{n+1}}, \quad n \ge 2. \qquad \textcircled{1}
Let mm be a positive integer and m≥2m \ge 2. Prove that for n≤mn \le m,
[an+32n+3]1m(m−(23)n(m−1)m)<m2−1m−n+1.2◯ \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}. \qquad \textcircled{2}

Solution

Proof By Equation (1), we have
2nan=3⋅2n−1an−1+34. 2^n a_n = 3 \cdot 2^{n-1} a_{n-1} + \frac{3}{4}.
Set bn=2nanb_n = 2^n a_n, n=1,2,…n = 1, 2, \dots, then
bn=3bn−1+34,bn+38=3(bn−1+38). b_n = 3b_{n-1} + \frac{3}{4}, \quad b_n + \frac{3}{8} = 3\left(b_{n-1} + \frac{3}{8}\right).
Since b1=2a1=218, \text{Since } b_1 = 2a_1 = \frac{21}{8},
bn+38=3n−1(b1+38)=3n, b_n + \frac{3}{8} = 3^{n-1} \left(b_1 + \frac{3}{8}\right) = 3^n,
it follows that
an=(32)n−32n+3 a_n = \left(\frac{3}{2}\right)^n - \frac{3}{2^{n+3}}

Therefore, in order to prove Equation ②, it suffices to prove that
(32)nm(m−(23)n(m−1)m)<m2−1m−n+1, \left(\frac{3}{2}\right)^{\frac{n}{m}} \left(m - \left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right) < \frac{m^2 - 1}{m - n + 1},
or equivalently,
(1−nm+1)(32)nm(m−(23)n(m−1)m)<m−1.3◯ \left(1 - \frac{n}{m+1}\right) \left(\frac{3}{2}\right)^{\frac{n}{m}} \left(m - \left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right) < m - 1. \quad \textcircled{3}

At first, we estimate the upper bound of 1−nm+11 - \frac{n}{m+1}. By using Bernoulli's inequality, we get
1−nm+1<(1−1m+1)n, 1 - \frac{n}{m+1} < \left(1 - \frac{1}{m+1}\right)^n,
so that
(1−nm+1)m<(1−1m+1)nm=(mm+1)nm=[1(1+1m)m]n. \left(1 - \frac{n}{m+1}\right)^m < \left(1 - \frac{1}{m+1}\right)^{nm} = \left(\frac{m}{m+1}\right)^{nm} = \left[ \frac{1}{\left(1 + \frac{1}{m}\right)^m} \right]^n.

(Note: By the mean inequality, we can also have the same result:
(1−nm+1)m=(1−nm+1)m⋅1⋅1⋅⋯⋅1⏟number mn−m of 1<[m(1−nm+1)+mn−mmn]mn=(mm+1)nm. \begin{align*} \left(1 - \frac{n}{m+1}\right)^m &= \left(1 - \frac{n}{m+1}\right)^m \cdot \underbrace{1 \cdot 1 \cdot \cdots \cdot 1}_{\text{number } mn-m \text{ of } 1} \\ &< \left[ \frac{m\left(1 - \frac{n}{m+1}\right) + mn - m}{mn} \right]^{mn} \\ &= \left(\frac{m}{m+1}\right)^{nm}. \end{align*}

Since m≥2m \ge 2, in view of the binomial formula, we obtain
(1+1mm)m≥1+Cm1⋅1m+Cm2⋅1m2=52−12m≥94. \left(1 + \frac{1}{mm}\right)^m \ge 1 + C_m^1 \cdot \frac{1}{m} + C_m^2 \cdot \frac{1}{m^2} = \frac{5}{2} - \frac{1}{2m} \ge \frac{9}{4}.
It follows that
(1−nm+1)m<(49)n, \left(1 - \frac{n}{m+1}\right)^m < \left(\frac{4}{9}\right)^n,
or
1−nm+1<(23)2nm. 1 - \frac{n}{m+1} < \left(\frac{2}{3}\right)^{\frac{2n}{m}}.
Hence, if we want to prove Equation ③, we only need to prove that
(23)2nm⋅(32)nm(m−(23)n(m−1)m)<m−1, \left(\frac{2}{3}\right)^{\frac{2n}{m}} \cdot \left(\frac{3}{2}\right)^{\frac{n}{m}} \left(m - \left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right) < m - 1,
that is,
(23)nm(m−(23)n(m−1)m)<m−1.4◯ \left(\frac{2}{3}\right)^{\frac{n}{m}} \left(m - \left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right) < m - 1. \quad \textcircled{4}
Set (23)nm=t\left(\frac{2}{3}\right)^{\frac{n}{m}} = t, then 0<t<10 < t < 1, and Equation ④ now becomes
t(m−tm−1)<m−1, t(m - t^{m-1}) < m - 1,
or
(t−1)[m−(tm−1+tm−2+⋯+1)]<0. (t-1)[m - (t^{m-1} + t^{m-2} + \cdots + 1)] < 0.
The above inequality clearly holds, so does the initial inequality.

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