China Mathematical Olympiad

Proof By Equation (1), we have $$2^n{a}_n=3\cdot 2^{n-1}{a}_{n-1}+\frac{3}{4}.$$ Set $b_n=2^n{a}_n$, $n=1,2,\dots$, then $$b_n=3b_{n-1}+\frac{3}{4},b_n+\frac{3}{8}=3\left(b_{n-1}+\frac{3}{8}\right).$$ $$\text{Since }{b}_1=2a_1=\frac{21}{8},$$ $$b_n+\frac{3}{8}=3^{n-1}\left(b_1+\frac{3}{8}\right)=3^n,$$ it follows that $$a_n={\left(\frac{3}{2}\right)}^n-\frac{3}{2^{n+3}}$$

Therefore, in order to prove Equation ②, it suffices to prove that $${\left(\frac{3}{2}\right)}^{\frac{n}{m}}\left(m-{\left(\frac{2}{3}\right)}^{\frac{n(m-1)}{m}}\right)<\frac{m^2-1}{m-n+1},$$ or equivalently, $$\left(1-\frac{n}{m+1}\right){\left(\frac{3}{2}\right)}^{\frac{n}{m}}\left(m-{\left(\frac{2}{3}\right)}^{\frac{n(m-1)}{m}}\right)<m-1.\stackrel{◯}{3}$$

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

(Note: By the mean inequality, we can also have the same result: $$\begin{array}{rl}{\left(1-\frac{n}{m+1}\right)}^m& ={\left(1-\frac{n}{m+1}\right)}^m\cdot \underset{\text{number }mn-m\text{ of }1}{\underbrace{1\cdot 1\cdot \cdots \cdot 1}}\\ & <{\left[\frac{m\left(1-\frac{n}{m+1}\right)+mn-m}{mn}\right]}^{mn}\\ & ={\left(\frac{m}{m+1}\right)}^{nm}.\end{array}$$

Since $m\ge 2$, in view of the binomial formula, we obtain $${\left(1+\frac{1}{mm}\right)}^m\ge 1+C_m^1\cdot \frac{1}{m}+C_m^2\cdot \frac{1}{m^2}=\frac{5}{2}-\frac{1}{2m}\ge \frac{9}{4}.$$ It follows that $${\left(1-\frac{n}{m+1}\right)}^m<{\left(\frac{4}{9}\right)}^n,$$ or $$1-\frac{n}{m+1}<{\left(\frac{2}{3}\right)}^{\frac{2n}{m}}.$$ Hence, if we want to prove Equation ③, we only need to prove that $${\left(\frac{2}{3}\right)}^{\frac{2n}{m}}\cdot {\left(\frac{3}{2}\right)}^{\frac{n}{m}}\left(m-{\left(\frac{2}{3}\right)}^{\frac{n(m-1)}{m}}\right)<m-1,$$ that is, $${\left(\frac{2}{3}\right)}^{\frac{n}{m}}\left(m-{\left(\frac{2}{3}\right)}^{\frac{n(m-1)}{m}}\right)<m-1.\stackrel{◯}{4}$$ Set ${\left(\frac{2}{3}\right)}^{\frac{n}{m}}=t$, then $0<t<1$, and Equation ④ now becomes $$t(m-t^{m-1})<m-1,$$ or $$(t-1)[m-(t^{m-1}+t^{m-2}+\cdots +1)]<0.$$ The above inequality clearly holds, so does the initial inequality.