Question about an inequation described by matrices
$A=(a_ij)_1 le i, j le n$ is a matrix that$sum_limitsi=1^n a_ij=1$ for every j and $sum_limitsj=1^n a_ij = 1$ for every i and $a_ij ge 0$.And
$$beginequation
beginpmatrix
y_1 \
vdots \
y_n \
endpmatrix
=mathbfA
beginpmatrix
x_1 \
vdots \
x_n
endpmatrix
endequation$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$
It may somehow matter to convex function.
linear-algebra inequalities convex-analysis
edited 11 hours ago
user44191
2,5431126
asked 12 hours ago
X.T Chen
112
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$A=(a_ij)_1 le i, j le n$ is a matrix that$sum_limitsi=1^n a_ij=1$ for every j and $sum_limitsj=1^n a_ij = 1$ for every i and $a_ij ge 0$.And
$$beginequation
beginpmatrix
y_1 \
vdots \
y_n \
endpmatrix
=mathbfA
beginpmatrix
x_1 \
vdots \
x_n
endpmatrix
endequation$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$
It may somehow matter to convex function.
linear-algebra inequalities convex-analysis
edited 11 hours ago
user44191
2,5431126
asked 12 hours ago
X.T Chen
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
$A=(a_ij)_1 le i, j le n$ is a matrix that$sum_limitsi=1^n a_ij=1$ for every j and $sum_limitsj=1^n a_ij = 1$ for every i and $a_ij ge 0$.And
$$beginequation
beginpmatrix
y_1 \
vdots \
y_n \
endpmatrix
=mathbfA
beginpmatrix
x_1 \
vdots \
x_n
endpmatrix
endequation$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$
It may somehow matter to convex function.
linear-algebra inequalities convex-analysis
edited 11 hours ago
user44191
2,5431126
asked 12 hours ago
X.T Chen
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$A=(a_ij)_1 le i, j le n$ is a matrix that$sum_limitsi=1^n a_ij=1$ for every j and $sum_limitsj=1^n a_ij = 1$ for every i and $a_ij ge 0$.And
$$beginequation
beginpmatrix
y_1 \
vdots \
y_n \
endpmatrix
=mathbfA
beginpmatrix
x_1 \
vdots \
x_n
endpmatrix
endequation$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$
It may somehow matter to convex function.
linear-algebra inequalities convex-analysis
linear-algebra inequalities convex-analysis
edited 11 hours ago
user44191
2,5431126
asked 12 hours ago
X.T Chen
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 11 hours ago
user44191
2,5431126
asked 12 hours ago
X.T Chen
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 11 hours ago
user44191
2,5431126
edited 11 hours ago
user44191
2,5431126
edited 11 hours ago
user44191
2,5431126
2,5431126
asked 12 hours ago
X.T Chen
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 12 hours ago
X.T Chen
112
asked 12 hours ago
X.T Chen
112
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
1 Answer
1
active
oldest
votes
By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,dots,y_n]^T$ and $x=[x,dots,x_n]^T$ is equivalent to the condition that $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.
answered 12 hours ago
Iosif Pinelis
17.7k12158
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,dots,y_n]^T$ and $x=[x,dots,x_n]^T$ is equivalent to the condition that $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.
answered 12 hours ago
Iosif Pinelis
17.7k12158
add a comment |Â
By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,dots,y_n]^T$ and $x=[x,dots,x_n]^T$ is equivalent to the condition that $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.
answered 12 hours ago
Iosif Pinelis
17.7k12158
add a comment |Â
By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,dots,y_n]^T$ and $x=[x,dots,x_n]^T$ is equivalent to the condition that $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.
answered 12 hours ago
Iosif Pinelis
17.7k12158
By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,dots,y_n]^T$ and $x=[x,dots,x_n]^T$ is equivalent to the condition that $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.
answered 12 hours ago
Iosif Pinelis
17.7k12158
answered 12 hours ago
Iosif Pinelis
17.7k12158
answered 12 hours ago
Iosif Pinelis
17.7k12158
answered 12 hours ago
Iosif Pinelis
17.7k12158
17.7k12158
add a comment |Â
add a comment |Â
X.T Chen is a new contributor. Be nice, and check out our Code of Conduct.
X.T Chen is a new contributor. Be nice, and check out our Code of Conduct.
X.T Chen is a new contributor. Be nice, and check out our Code of Conduct.
X.T Chen is a new contributor. Be nice, and check out our Code of Conduct.
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