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$prod_{i=1}^{n}x_i^{p_i}le p_1x_1+ldots +p_nx_n$ strict inequality when $p_i neq p_j$ for $i neq j$
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Let $x_i,p_i in mathbb{R}$ and $x_i,p_i>0$. I just showed that for $sum_{i=1}^{n}p_i=1$ we have the following inequality:
$$prod_{i=1}^{n}x_i^{p_i}le p_1x_1+ldots +p_nx_n$$
(I used Jensen's inequality for this)
Now the next question is why we do we have the strict inequality if all $x_i$ are different:
$$prod_{i=1}^{n}x_i^{p_i}< p_1x_1+ldots +p_nx_n$$ for $x_ineq x_j$for $i neq j$.
Can somebody answer this? Thanks in advance!
real-analysis inequality
asked yesterday
user610431
627
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up vote
1
down vote
favorite
Let $x_i,p_i in mathbb{R}$ and $x_i,p_i>0$. I just showed that for $sum_{i=1}^{n}p_i=1$ we have the following inequality:
$$prod_{i=1}^{n}x_i^{p_i}le p_1x_1+ldots +p_nx_n$$
(I used Jensen's inequality for this)
Now the next question is why we do we have the strict inequality if all $x_i$ are different:
$$prod_{i=1}^{n}x_i^{p_i}< p_1x_1+ldots +p_nx_n$$ for $x_ineq x_j$for $i neq j$.
Can somebody answer this? Thanks in advance!
real-analysis inequality
asked yesterday
user610431
627
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $x_i,p_i in mathbb{R}$ and $x_i,p_i>0$. I just showed that for $sum_{i=1}^{n}p_i=1$ we have the following inequality:
$$prod_{i=1}^{n}x_i^{p_i}le p_1x_1+ldots +p_nx_n$$
(I used Jensen's inequality for this)
Now the next question is why we do we have the strict inequality if all $x_i$ are different:
$$prod_{i=1}^{n}x_i^{p_i}< p_1x_1+ldots +p_nx_n$$ for $x_ineq x_j$for $i neq j$.
Can somebody answer this? Thanks in advance!
real-analysis inequality
asked yesterday
user610431
627
Let $x_i,p_i in mathbb{R}$ and $x_i,p_i>0$. I just showed that for $sum_{i=1}^{n}p_i=1$ we have the following inequality:
$$prod_{i=1}^{n}x_i^{p_i}le p_1x_1+ldots +p_nx_n$$
(I used Jensen's inequality for this)
Now the next question is why we do we have the strict inequality if all $x_i$ are different:
$$prod_{i=1}^{n}x_i^{p_i}< p_1x_1+ldots +p_nx_n$$ for $x_ineq x_j$for $i neq j$.
Can somebody answer this? Thanks in advance!
real-analysis inequality
real-analysis inequality
asked yesterday
user610431
627
asked yesterday
user610431
627
asked yesterday
user610431
627
asked yesterday
user610431
627
asked yesterday
user610431
627
627
add a comment |
add a comment |
1 Answer
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You mentioned that you used Jensen's inequality, which tells you that
$$prod_{i=1}^n x_i^{p_i} = mathrm{exp} left( sum_{i=1}^n p_i log x_i right) le sum_{i=1}^n p_i , mathrm{exp}(log x_i) = sum_{i=1}^n p_ix_i$$
since $x mapsto mathrm{exp}(x)$ is convex.
But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $log x_1 = log x_2 = cdots = log x_n$. But this in turn holds if and only if $x_1 = x_2 = cdots = x_n$.
So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.
answered yesterday
Clive Newstead
49.2k472132
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You mentioned that you used Jensen's inequality, which tells you that
$$prod_{i=1}^n x_i^{p_i} = mathrm{exp} left( sum_{i=1}^n p_i log x_i right) le sum_{i=1}^n p_i , mathrm{exp}(log x_i) = sum_{i=1}^n p_ix_i$$
since $x mapsto mathrm{exp}(x)$ is convex.
But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $log x_1 = log x_2 = cdots = log x_n$. But this in turn holds if and only if $x_1 = x_2 = cdots = x_n$.
So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.
answered yesterday
Clive Newstead
49.2k472132
add a comment |
up vote
2
down vote
accepted
You mentioned that you used Jensen's inequality, which tells you that
$$prod_{i=1}^n x_i^{p_i} = mathrm{exp} left( sum_{i=1}^n p_i log x_i right) le sum_{i=1}^n p_i , mathrm{exp}(log x_i) = sum_{i=1}^n p_ix_i$$
since $x mapsto mathrm{exp}(x)$ is convex.
But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $log x_1 = log x_2 = cdots = log x_n$. But this in turn holds if and only if $x_1 = x_2 = cdots = x_n$.
So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.
answered yesterday
Clive Newstead
49.2k472132
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You mentioned that you used Jensen's inequality, which tells you that
$$prod_{i=1}^n x_i^{p_i} = mathrm{exp} left( sum_{i=1}^n p_i log x_i right) le sum_{i=1}^n p_i , mathrm{exp}(log x_i) = sum_{i=1}^n p_ix_i$$
since $x mapsto mathrm{exp}(x)$ is convex.
But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $log x_1 = log x_2 = cdots = log x_n$. But this in turn holds if and only if $x_1 = x_2 = cdots = x_n$.
So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.
answered yesterday
Clive Newstead
49.2k472132
You mentioned that you used Jensen's inequality, which tells you that
$$prod_{i=1}^n x_i^{p_i} = mathrm{exp} left( sum_{i=1}^n p_i log x_i right) le sum_{i=1}^n p_i , mathrm{exp}(log x_i) = sum_{i=1}^n p_ix_i$$
since $x mapsto mathrm{exp}(x)$ is convex.
But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $log x_1 = log x_2 = cdots = log x_n$. But this in turn holds if and only if $x_1 = x_2 = cdots = x_n$.
So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.
answered yesterday
Clive Newstead
49.2k472132
answered yesterday
Clive Newstead
49.2k472132
answered yesterday
Clive Newstead
49.2k472132
answered yesterday
Clive Newstead
49.2k472132
49.2k472132
add a comment |
add a comment |
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