why can we use this $x^theta y^{1-theta} le theta x+(1-theta)y$ to prove the $prod limits_{i=1}^{n}x_i ge 1$...












0














Show that ${x in R^n_+|prod limits_{i=1}^{n}x_i ge 1}$ is convex set



Hint : if $x ,y ge 0$ and $0le theta le 1$,then $x^theta y^{1-theta} le theta x+(1-theta)y$



I don't understand the relation between the problem and the hint,the question wants me to prove the $prod limits_{i=1}^{n}x_i ge 1$ is a convex,why shouldn't we use the definition of convex: $f(theta x+(1-theta )y) le theta f(x)+(1-theta)f(y)$,but use this $x^theta y^{1-theta} le theta x+(1-theta)y$ to prove the $prod limits_{i=1}^{n}x_i ge 1$ is convex set?






convex-analysis convex-optimization







share|cite|improve this question


















  • 2




    Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
    – Michal Adamaszek
    Nov 30 at 13:54










  • I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
    – DonAntonio
    Nov 30 at 14:03
















0














Show that ${x in R^n_+|prod limits_{i=1}^{n}x_i ge 1}$ is convex set



Hint : if $x ,y ge 0$ and $0le theta le 1$,then $x^theta y^{1-theta} le theta x+(1-theta)y$



I don't understand the relation between the problem and the hint,the question wants me to prove the $prod limits_{i=1}^{n}x_i ge 1$ is a convex,why shouldn't we use the definition of convex: $f(theta x+(1-theta )y) le theta f(x)+(1-theta)f(y)$,but use this $x^theta y^{1-theta} le theta x+(1-theta)y$ to prove the $prod limits_{i=1}^{n}x_i ge 1$ is convex set?






convex-analysis convex-optimization







share|cite|improve this question


















  • 2




    Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
    – Michal Adamaszek
    Nov 30 at 13:54










  • I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
    – DonAntonio
    Nov 30 at 14:03














0












0








0







Show that ${x in R^n_+|prod limits_{i=1}^{n}x_i ge 1}$ is convex set



Hint : if $x ,y ge 0$ and $0le theta le 1$,then $x^theta y^{1-theta} le theta x+(1-theta)y$



I don't understand the relation between the problem and the hint,the question wants me to prove the $prod limits_{i=1}^{n}x_i ge 1$ is a convex,why shouldn't we use the definition of convex: $f(theta x+(1-theta )y) le theta f(x)+(1-theta)f(y)$,but use this $x^theta y^{1-theta} le theta x+(1-theta)y$ to prove the $prod limits_{i=1}^{n}x_i ge 1$ is convex set?






convex-analysis convex-optimization







share|cite|improve this question













Show that ${x in R^n_+|prod limits_{i=1}^{n}x_i ge 1}$ is convex set



Hint : if $x ,y ge 0$ and $0le theta le 1$,then $x^theta y^{1-theta} le theta x+(1-theta)y$



I don't understand the relation between the problem and the hint,the question wants me to prove the $prod limits_{i=1}^{n}x_i ge 1$ is a convex,why shouldn't we use the definition of convex: $f(theta x+(1-theta )y) le theta f(x)+(1-theta)f(y)$,but use this $x^theta y^{1-theta} le theta x+(1-theta)y$ to prove the $prod limits_{i=1}^{n}x_i ge 1$ is convex set?






convex-analysis convex-optimization




convex-analysis convex-optimization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 at 13:50









shineele

377




377








  • 2




    Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
    – Michal Adamaszek
    Nov 30 at 13:54










  • I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
    – DonAntonio
    Nov 30 at 14:03














  • 2




    Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
    – Michal Adamaszek
    Nov 30 at 13:54










  • I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
    – DonAntonio
    Nov 30 at 14:03








2




2




Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
– Michal Adamaszek
Nov 30 at 13:54




Start with the definition of a convex set (not function) and carry out the rest of your plan. Then you will see when to use the hint.
– Michal Adamaszek
Nov 30 at 13:54












I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
– DonAntonio
Nov 30 at 14:03




I think that carrying on the usual definition is pretty straightforward. I also can't see what that hint is useful for...not even if it simplifies something the proof...
– DonAntonio
Nov 30 at 14:03










1 Answer
1






active

oldest

votes


















2














you may want to use this hint the other way: from right to left...



let's call $mathbb{A}$ your ensemble, let's take $(x,y) in mathbb{A}^2$. to prove that $mathbb{A}$ is convex, we need to prove that $forall t in [0,1], z = tx+(1-t)y ;$ is in $mathbb{A}$



$Pi_{i=1}^n z_i = Pi_{i=1}^n (tx_i+(1-t)y_i)$



now you use the inequality from the right side to the left.



$Pi_{i=1}^n (tx_i+(1-t)y_i) > Pi_{i=1}^n x_i^ty_i^{1-t}$



you are now one line away from the result






share|cite|improve this answer





















  • how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
    – shineele
    Nov 30 at 14:22










  • how i came to this formula
    – shineele
    Nov 30 at 14:26












  • it's explained in the answer...
    – Alexis
    Nov 30 at 14:27










  • i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
    – shineele
    Nov 30 at 14:29












  • for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
    – Alexis
    Nov 30 at 14:31











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1 Answer
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1 Answer
1






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active

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active

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2














you may want to use this hint the other way: from right to left...



let's call $mathbb{A}$ your ensemble, let's take $(x,y) in mathbb{A}^2$. to prove that $mathbb{A}$ is convex, we need to prove that $forall t in [0,1], z = tx+(1-t)y ;$ is in $mathbb{A}$



$Pi_{i=1}^n z_i = Pi_{i=1}^n (tx_i+(1-t)y_i)$



now you use the inequality from the right side to the left.



$Pi_{i=1}^n (tx_i+(1-t)y_i) > Pi_{i=1}^n x_i^ty_i^{1-t}$



you are now one line away from the result






share|cite|improve this answer





















  • how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
    – shineele
    Nov 30 at 14:22










  • how i came to this formula
    – shineele
    Nov 30 at 14:26












  • it's explained in the answer...
    – Alexis
    Nov 30 at 14:27










  • i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
    – shineele
    Nov 30 at 14:29












  • for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
    – Alexis
    Nov 30 at 14:31
















2














you may want to use this hint the other way: from right to left...



let's call $mathbb{A}$ your ensemble, let's take $(x,y) in mathbb{A}^2$. to prove that $mathbb{A}$ is convex, we need to prove that $forall t in [0,1], z = tx+(1-t)y ;$ is in $mathbb{A}$



$Pi_{i=1}^n z_i = Pi_{i=1}^n (tx_i+(1-t)y_i)$



now you use the inequality from the right side to the left.



$Pi_{i=1}^n (tx_i+(1-t)y_i) > Pi_{i=1}^n x_i^ty_i^{1-t}$



you are now one line away from the result






share|cite|improve this answer





















  • how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
    – shineele
    Nov 30 at 14:22










  • how i came to this formula
    – shineele
    Nov 30 at 14:26












  • it's explained in the answer...
    – Alexis
    Nov 30 at 14:27










  • i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
    – shineele
    Nov 30 at 14:29












  • for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
    – Alexis
    Nov 30 at 14:31














2












2








2






you may want to use this hint the other way: from right to left...



let's call $mathbb{A}$ your ensemble, let's take $(x,y) in mathbb{A}^2$. to prove that $mathbb{A}$ is convex, we need to prove that $forall t in [0,1], z = tx+(1-t)y ;$ is in $mathbb{A}$



$Pi_{i=1}^n z_i = Pi_{i=1}^n (tx_i+(1-t)y_i)$



now you use the inequality from the right side to the left.



$Pi_{i=1}^n (tx_i+(1-t)y_i) > Pi_{i=1}^n x_i^ty_i^{1-t}$



you are now one line away from the result






share|cite|improve this answer












you may want to use this hint the other way: from right to left...



let's call $mathbb{A}$ your ensemble, let's take $(x,y) in mathbb{A}^2$. to prove that $mathbb{A}$ is convex, we need to prove that $forall t in [0,1], z = tx+(1-t)y ;$ is in $mathbb{A}$



$Pi_{i=1}^n z_i = Pi_{i=1}^n (tx_i+(1-t)y_i)$



now you use the inequality from the right side to the left.



$Pi_{i=1}^n (tx_i+(1-t)y_i) > Pi_{i=1}^n x_i^ty_i^{1-t}$



you are now one line away from the result







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 at 14:05









Alexis

2199




2199












  • how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
    – shineele
    Nov 30 at 14:22










  • how i came to this formula
    – shineele
    Nov 30 at 14:26












  • it's explained in the answer...
    – Alexis
    Nov 30 at 14:27










  • i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
    – shineele
    Nov 30 at 14:29












  • for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
    – Alexis
    Nov 30 at 14:31


















  • how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
    – shineele
    Nov 30 at 14:22










  • how i came to this formula
    – shineele
    Nov 30 at 14:26












  • it's explained in the answer...
    – Alexis
    Nov 30 at 14:27










  • i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
    – shineele
    Nov 30 at 14:29












  • for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
    – Alexis
    Nov 30 at 14:31
















how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
– shineele
Nov 30 at 14:22




how does $Pi_{i=1}^n x_i^ty_i^{1-t}$ come ?
– shineele
Nov 30 at 14:22












how i came to this formula
– shineele
Nov 30 at 14:26






how i came to this formula
– shineele
Nov 30 at 14:26














it's explained in the answer...
– Alexis
Nov 30 at 14:27




it's explained in the answer...
– Alexis
Nov 30 at 14:27












i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
– shineele
Nov 30 at 14:29






i mean why isn't $Pi_{i=1}^n (tx_i+(1-t)y_i) > 0 $ or the other formula or number,why must we $> Pi_{i=1}^n x_i^ty_i^{1-t}$?
– shineele
Nov 30 at 14:29














for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
– Alexis
Nov 30 at 14:31




for two reasons: - one it's the hint, you were looking for a way to use the hint. here it is. if you change this you don't use the hint. - it brings you very very close to the results and it's after that straight forward to prove that z is in $mathbb{A}$
– Alexis
Nov 30 at 14:31


















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