Let the sequence $(a_{n})_{n\geqslant 1}$ be defined as: $$a_{n}=\sqrt{A_{n+2}^{1}\sqrt[3]{A_{n+3}^{2}\sqrt[4]{A_{n+4}^{3}\sqrt[5]{A_{n+5}^{4}}}}},$$where $A_{m}^{k}$ are defined by $$A_{m}^{k}=\binom{m}{k}\cdot k!.$$Prove that $$a_{n}<\frac{119}{120}\cdot n+\frac{7}{3}.$$
Problem
Source: Moldova TST 2017, Day 1, Problem 1
Tags: algebra
---Fermat---
06.03.2017 23:13
This problem is immediately killed by A.M-G.M. inequality.
\begin{align*}
a_{n}&=\sqrt{A_{n+2}^{1}\sqrt[3]{A_{n+3}^{2}\sqrt[4]{A_{n+4}^{3}\sqrt[5]{A_{n+5}^{4}}}}} \\
&=\sqrt{(n+2)\sqrt[3]{(n+2)(n+3)\sqrt[4]{(n+2)(n+3)(n+4)\sqrt[5]{(n+2)(n+3)(n+4)(n+5)}}}} \\
&=\sqrt[120]{(n+2)^{86}(n+3)^{26}(n+4)^5(n+5)} \\
&\leq \frac{86(n+2)+26(n+3)+5(n+4)+(n+5)+1+1}{120} \\
&= \frac{118n+277}{120} \\
&< \frac{119}{120} \cdot n + \frac{7}{3}.
\end{align*}P.S Please tell me if there is something wrong.
rmtf1111
06.03.2017 23:15
@above, yes it is right, and the problem is indeed easy, it was solved by everyone , except of one contestant
seoneo
15.10.2017 05:57
A slight correction for the above;
\begin{align*}
a_{n}&=\sqrt{A_{n+2}^{1}\sqrt[3]{A_{n+3}^{2}\sqrt[4]{A_{n+4}^{3}\sqrt[5]{A_{n+5}^{4}}}}} \\
&=\sqrt{(n+2)\sqrt[3]{(n+2)(n+3)\sqrt[4]{(n+2)(n+3)(n+4)\sqrt[5]{(n+2)(n+3)(n+4)(n+5)}}}} \\
&=\sqrt[120]{(n+2)^{86}(n+3)^{26}(n+4)^6(n+5)} \\
&\leq \frac{86(n+2)+26(n+3)+6(n+4)+(n+5)+1}{120} \\
&= \frac{119}{120} \cdot n + \frac{7}{3}.
\end{align*}and no equality is possible.
sqing
25.01.2018 10:54
Snakes wrote: Let the sequence $(a_{n})_{n\geqslant 1}$ be defined as: $$a_{n}=\sqrt{A_{n+2}^{1}\sqrt[3]{A_{n+3}^{2}\sqrt[4]{A_{n+4}^{3}\sqrt[5]{A_{n+5}^{4}}}}}$$Where $A_{m}^{k}$ are arrangements: $A_{m}^{k}=\binom{m}{k}\cdot k!$ Prove that $a_{n}<\frac{119}{120}\cdot n+\frac{7}{3}$ Let the sequence $(a_{n})_{n\geqslant 1}$ be defined as: $$a_{n}=\sqrt{A_{n+2}^{1}\sqrt[3]{A_{n+3}^{2}\sqrt[4]{A_{n+4}^{3}\sqrt[5]{A_{n+5}^{4}}}}}$$Where $A_{m}^{k}$ are arrangements: $A_{m}^{k}=\binom{m}{k}\cdot k!$ Prove that $$a_{n}<n^{\frac{119}{120}}+\frac{93}{40}$$