Difference between revisions of "2020 INMO Problems/Problem 4"
(Created page with "==Problem== Let <math>n \geqslant 2</math> be an integer and let <math>1<a_1 \le a_2 \le \dots \le a_n</math> be <math>n</math> real numbers such that <math>a_1+a_2+\dots+a_n...") |
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==Solution(2)== | ==Solution(2)== | ||
− | <math>\boxed{\text{Notation}}</math> | + | <math>\boxed{\text{Notation}}</math> |
Define,<math>S_1=a_1+a_1a_2+\cdots +a_1\cdots a_{n-1}+2</math>. | Define,<math>S_1=a_1+a_1a_2+\cdots +a_1\cdots a_{n-1}+2</math>. | ||
Line 30: | Line 30: | ||
---------- | ---------- | ||
− | + | <math>\boxed{\text{Claim(1)}}</math> | |
<math>S_1 <2^{n-1}</math>. | <math>S_1 <2^{n-1}</math>. | ||
− | + | <math>\textbf{ Proof:}</math> | |
Using Tchevbycev inequality we have , | Using Tchevbycev inequality we have , | ||
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<math><2^{n-1}</math>. | <math><2^{n-1}</math>. | ||
− | + | Since,<math>a_1+\cdots +a_{n} =2n</math> and <math>a_n\ge 2</math> , Hence ,<math>a_1+\cdots +a_{n-1}\le 2n-2</math> . | |
------------- | ------------- | ||
− | + | <math>\boxed{\text{Claim(2)}}</math> | |
<math>2^{n-1}< a_1\cdots a_n \le 2^{n}</math>. | <math>2^{n-1}< a_1\cdots a_n \le 2^{n}</math>. | ||
− | + | <math>\textbf{Proof}</math> | |
The RHS inequality is trivial by AM-GM inequality. | The RHS inequality is trivial by AM-GM inequality. | ||
Line 84: | Line 84: | ||
------------- | ------------- | ||
− | 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>. | + | 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 be an integer and let
be
real numbers such that
. Prove that
Solution(1)
For , we want to show that
where
and
. This is equivalent to showing that
, which is true.
Suppose, now, that the given inequality is true for , where
. Now, consider
reals
with sum
. Then,
and
, so by induction hypothesis,
This means
or
as desired. ~biomathematics
Solution(2)
Define,.
,
In general ,.
.
.
Using Tchevbycev inequality we have ,
.
.
.
.[Applying Induction on successive
].
.
.
.
.[using GM-AM]
.
Since, and
, Hence ,
.
.
The RHS inequality is trivial by AM-GM inequality.
For LHS inequality I would like to use induction.
.
We have ,
and
.
.
Suppose , the statement is true for
such that ,
and
.
Now , consider .
Suppose , is median of the sequence,
and
and
.
and
.
Our induction step is complete.
This two claim leads and equality for
. ~trishan