Inequality Question - Homogeneity
$begingroup$
Let $n>3$ and $x_1, x_2, ldots, x_n$ be positive real numbers with $x_1x_2cdots x_n=1$. Prove that
$$
frac{1}{1+x_1+x_1x_2}+frac{1}{1+x_2+x_2x_3}+cdots+frac{1}{1+x_n+x_nx_1}>1.
$$
I’d like to homogenize the degree of the denominator of each term but I ran out of tricks. Any help will be appreciated. Thanks.
inequality summation substitution
edited Dec 20 '18 at 4:34
Michael Rozenberg
108k1895200
asked Dec 20 '18 at 3:13
BBTBBT
113
$endgroup$
add a comment |
$begingroup$
Let $n>3$ and $x_1, x_2, ldots, x_n$ be positive real numbers with $x_1x_2cdots x_n=1$. Prove that
$$
frac{1}{1+x_1+x_1x_2}+frac{1}{1+x_2+x_2x_3}+cdots+frac{1}{1+x_n+x_nx_1}>1.
$$
I’d like to homogenize the degree of the denominator of each term but I ran out of tricks. Any help will be appreciated. Thanks.
inequality summation substitution
edited Dec 20 '18 at 4:34
Michael Rozenberg
108k1895200
asked Dec 20 '18 at 3:13
BBTBBT
113
$endgroup$
1
$begingroup$
Which tricks have you tried?
$endgroup$
– Alexander Burstein
Dec 20 '18 at 3:48
add a comment |
$begingroup$
Let $n>3$ and $x_1, x_2, ldots, x_n$ be positive real numbers with $x_1x_2cdots x_n=1$. Prove that
$$
frac{1}{1+x_1+x_1x_2}+frac{1}{1+x_2+x_2x_3}+cdots+frac{1}{1+x_n+x_nx_1}>1.
$$
I’d like to homogenize the degree of the denominator of each term but I ran out of tricks. Any help will be appreciated. Thanks.
inequality summation substitution
edited Dec 20 '18 at 4:34
Michael Rozenberg
108k1895200
asked Dec 20 '18 at 3:13
BBTBBT
113
$endgroup$
Let $n>3$ and $x_1, x_2, ldots, x_n$ be positive real numbers with $x_1x_2cdots x_n=1$. Prove that
$$
frac{1}{1+x_1+x_1x_2}+frac{1}{1+x_2+x_2x_3}+cdots+frac{1}{1+x_n+x_nx_1}>1.
$$
I’d like to homogenize the degree of the denominator of each term but I ran out of tricks. Any help will be appreciated. Thanks.
inequality summation substitution
inequality summation substitution
edited Dec 20 '18 at 4:34
Michael Rozenberg
108k1895200
asked Dec 20 '18 at 3:13
BBTBBT
113
edited Dec 20 '18 at 4:34
Michael Rozenberg
108k1895200
asked Dec 20 '18 at 3:13
BBTBBT
113
edited Dec 20 '18 at 4:34
Michael Rozenberg
108k1895200
edited Dec 20 '18 at 4:34
Michael Rozenberg
108k1895200
edited Dec 20 '18 at 4:34
Michael Rozenberg
108k1895200
108k1895200
asked Dec 20 '18 at 3:13
BBTBBT
113
asked Dec 20 '18 at 3:13
BBTBBT
113
asked Dec 20 '18 at 3:13
BBTBBT
113
113
1
$begingroup$
Which tricks have you tried?
$endgroup$
– Alexander Burstein
Dec 20 '18 at 3:48
add a comment |
1
$begingroup$
Which tricks have you tried?
$endgroup$
– Alexander Burstein
Dec 20 '18 at 3:48
1
1
$begingroup$
Which tricks have you tried?
$endgroup$
– Alexander Burstein
Dec 20 '18 at 3:48
$begingroup$
Which tricks have you tried?
$endgroup$
– Alexander Burstein
Dec 20 '18 at 3:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can make the homogenization by the following way.
Let $x_1=frac{a_2}{a_1},$ $x_2=frac{a_3}{a_2},$..., $x_{n-1}=frac{a_n}{a_{n-1}},$ where all $a_i>0$, $a_{n+1}=a_1$ and $a_{n+2}=a_2$.
Thus, the condition gives $x_n=frac{a_1}{a_n}$ and
$$sum_{i=1}^nfrac{1}{1+x_1+x_2}=sum_{i=1}^nfrac{1}{1+frac{a_{i+1}}{a_i}+frac{a_{i+2}}{a_i}}=sum_{i=1}^nfrac{a_i}{a_i+a_{i+1}+a_{i+2}}>sum_{i=1}^nfrac{a_i}{a_1+a_2+...+a_n}=1.$$
answered Dec 20 '18 at 4:27
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
This is great. Thanks so much.
$endgroup$
– BBT
Dec 21 '18 at 0:02
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can make the homogenization by the following way.
Let $x_1=frac{a_2}{a_1},$ $x_2=frac{a_3}{a_2},$..., $x_{n-1}=frac{a_n}{a_{n-1}},$ where all $a_i>0$, $a_{n+1}=a_1$ and $a_{n+2}=a_2$.
Thus, the condition gives $x_n=frac{a_1}{a_n}$ and
$$sum_{i=1}^nfrac{1}{1+x_1+x_2}=sum_{i=1}^nfrac{1}{1+frac{a_{i+1}}{a_i}+frac{a_{i+2}}{a_i}}=sum_{i=1}^nfrac{a_i}{a_i+a_{i+1}+a_{i+2}}>sum_{i=1}^nfrac{a_i}{a_1+a_2+...+a_n}=1.$$
answered Dec 20 '18 at 4:27
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
This is great. Thanks so much.
$endgroup$
– BBT
Dec 21 '18 at 0:02
add a comment |
$begingroup$
You can make the homogenization by the following way.
Let $x_1=frac{a_2}{a_1},$ $x_2=frac{a_3}{a_2},$..., $x_{n-1}=frac{a_n}{a_{n-1}},$ where all $a_i>0$, $a_{n+1}=a_1$ and $a_{n+2}=a_2$.
Thus, the condition gives $x_n=frac{a_1}{a_n}$ and
$$sum_{i=1}^nfrac{1}{1+x_1+x_2}=sum_{i=1}^nfrac{1}{1+frac{a_{i+1}}{a_i}+frac{a_{i+2}}{a_i}}=sum_{i=1}^nfrac{a_i}{a_i+a_{i+1}+a_{i+2}}>sum_{i=1}^nfrac{a_i}{a_1+a_2+...+a_n}=1.$$
answered Dec 20 '18 at 4:27
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
This is great. Thanks so much.
$endgroup$
– BBT
Dec 21 '18 at 0:02
add a comment |
$begingroup$
You can make the homogenization by the following way.
Let $x_1=frac{a_2}{a_1},$ $x_2=frac{a_3}{a_2},$..., $x_{n-1}=frac{a_n}{a_{n-1}},$ where all $a_i>0$, $a_{n+1}=a_1$ and $a_{n+2}=a_2$.
Thus, the condition gives $x_n=frac{a_1}{a_n}$ and
$$sum_{i=1}^nfrac{1}{1+x_1+x_2}=sum_{i=1}^nfrac{1}{1+frac{a_{i+1}}{a_i}+frac{a_{i+2}}{a_i}}=sum_{i=1}^nfrac{a_i}{a_i+a_{i+1}+a_{i+2}}>sum_{i=1}^nfrac{a_i}{a_1+a_2+...+a_n}=1.$$
answered Dec 20 '18 at 4:27
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
You can make the homogenization by the following way.
Let $x_1=frac{a_2}{a_1},$ $x_2=frac{a_3}{a_2},$..., $x_{n-1}=frac{a_n}{a_{n-1}},$ where all $a_i>0$, $a_{n+1}=a_1$ and $a_{n+2}=a_2$.
Thus, the condition gives $x_n=frac{a_1}{a_n}$ and
$$sum_{i=1}^nfrac{1}{1+x_1+x_2}=sum_{i=1}^nfrac{1}{1+frac{a_{i+1}}{a_i}+frac{a_{i+2}}{a_i}}=sum_{i=1}^nfrac{a_i}{a_i+a_{i+1}+a_{i+2}}>sum_{i=1}^nfrac{a_i}{a_1+a_2+...+a_n}=1.$$
answered Dec 20 '18 at 4:27
Michael RozenbergMichael Rozenberg
108k1895200
edited Dec 20 '18 at 9:16
answered Dec 20 '18 at 4:27
Michael RozenbergMichael Rozenberg
108k1895200
answered Dec 20 '18 at 4:27
Michael RozenbergMichael Rozenberg
108k1895200
answered Dec 20 '18 at 4:27
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
$begingroup$
This is great. Thanks so much.
$endgroup$
– BBT
Dec 21 '18 at 0:02
add a comment |
$begingroup$
This is great. Thanks so much.
$endgroup$
– BBT
Dec 21 '18 at 0:02
$begingroup$
This is great. Thanks so much.
$endgroup$
– BBT
Dec 21 '18 at 0:02
$begingroup$
This is great. Thanks so much.
$endgroup$
– BBT
Dec 21 '18 at 0:02
add a comment |
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$begingroup$
Which tricks have you tried?
$endgroup$
– Alexander Burstein
Dec 20 '18 at 3:48