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Difference between revisions of "2020 INMO Problems/Problem 4"

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Define,<math>S_1=a_1+a_1a_2+\cdots +a_1\cdots a_{n-1}+2</math>.
 
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This two claim leads<math>S_1<a_i \cdots a_n</math> and equality for <math>a_1=a_2=\cdots =a_n=2</math>.
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This two claim leads<math>S_1<a_i \cdots a_n</math> and equality for <math>a_1=a_2=\cdots =a_n=2</math>. ~trishan

Latest revision as of 13:34, 5 November 2020

Problem

Let $n \geqslant 2$ be an integer and let $1<a_1 \le a_2 \le \dots \le a_n$ be $n$ real numbers such that $a_1+a_2+\dots+a_n=2n$. Prove that\[a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n-2}+\dots+a_1a_2+a_1+2 \leqslant a_1a_2\dots a_n.\]

Solution(1)

For $n=2$, we want to show that \[a_1 + 2 \le a_1a_2\] where $a_1+a_2 = 4$ and $1 < a_1 \le a_2$. This is equivalent to showing that $(a_1-1)(a_1-2) \le 0$, which is true.

Suppose, now, that the given inequality is true for $n=k$, where $k \ge 2$. Now, consider $k+1$ reals $1 < a_1 \le a_2 \le \cdots \le a_{k+1}$ with sum $2k+2$. Then, $1 < a_1 \le a_2 \le \cdots \le a_{k-1} \le a_k + a_{k+1}-2$ and $a_1 + a_2 + \cdots + a_{k-1}+(a_k+a_{k+1}-2) = 2k$, so by induction hypothesis,

\[\sum_{i=1}^{k-1} a_1a_2 \cdots a_i + 2 \le a_1a_2 \cdots a_{k-1} \cdot (a_k + a_{k+1} -2)\] This means \[\sum_{i=1}^k a_1a_2 \cdots a_i + 2 \le a_1a_2 \cdots a_{k-1} \cdot (2a_k + a_{k+1} - 2) = a_1a_2 \cdots a_{k-1} \cdot (a_ka_{k+1}-(a_k-1)(a_{k+1}-2)) \le a_1a_2 \cdots a_{k-1} \cdot (a_ka_{k+1})\] or \[a_1a_2 \cdots a_k + a_1a_2 \cdots a_{k-1} + \cdots + a_1a_2 + a_1 + 2 \leq a_1a_2 \cdots a_{k+1}\] as desired. ~biomathematics


Solution(2)

$\boxed{\text{Notation}}$

Define,$S_1=a_1+a_1a_2+\cdots +a_1\cdots a_{n-1}+2$. $S_2=a_2+a_2a_3+\cdots +a_2\cdots a_{n-1}+2$,

In general ,$S_i =a_i +\cdots + a_i\cdots a_{n-1}+2$.

$\forall i< n$.

$\boxed{\text{Claim(1)}}$

$S_1 <2^{n-1}$.

$\textbf{ Proof:}$

Using Tchevbycev inequality we have , $\frac{S_1}{n} \le \frac {(n-1)a_1+2}{n} \frac{S_2}{n}$.

$\implies S_1 \le \frac {(n-1)a_1+2}{n} S_2$.

$\le \frac {(n-1)a_1+2}{n} . \frac{ (n-2)a_2+2}{n-1} S_3$.

$\le \prod _{i=1}^{n+1} \frac {(n-i)a_i +2}{(n+1-i)}$.[Applying Induction on successive $S_i$].

$=\prod_{i=1}^{n-1} (a_i +\frac{2-a_i}{n+1-i})$.

$< \prod_{i=1}^{n-1} (a_i +\frac {2-a_i}{2}$.

$=\prod_{i=1}^{n-1} \frac {a_i+2}{2}$.

$\le ( \frac{\frac{a_1+\cdots +a_{n-1}}{2}+(n-1)}{n-1})^{n-1}$.[using GM-AM]


$<2^{n-1}$.

Since,$a_1+\cdots +a_{n} =2n$ and $a_n\ge 2$ , Hence ,$a_1+\cdots +a_{n-1}\le 2n-2$ .

$\boxed{\text{Claim(2)}}$ $2^{n-1}< a_1\cdots a_n \le 2^{n}$.

$\textbf{Proof}$ 

The RHS inequality is trivial by AM-GM inequality.

For LHS inequality I would like to use induction.

$\boxed{n=2}$.

We have $a_1+a_2=4$ ,$1<a_1\le 2$ and $a_2\ge 2$.

$\implies a_1a_2 > 2$. Suppose , the statement is true for$n=k$ such that , $a_1+\cdots +a_k =2k$ and $a_1 \cdots a_K>2^{k-1}$.

Now , consider $1<b_1 \le b_2 \cdots \le b_{k+1}$.

Suppose ,$b_j =2$ is median of the sequence, and$a_i=b_i ,\forall i<j$ and $a_i=b_{i+1},\forall k\ge i>j$.

$\implies b_1+...+b_{k+1}=2k+2$ and $b_1..b_{k+1}>2^k$.

Our induction step is complete.

This two claim leads$S_1<a_i \cdots a_n$ and equality for $a_1=a_2=\cdots =a_n=2$. ~trishan

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