Mixing
Proving $sum_{i=1}^{n}frac{1}{p^{a_i}} geq sum_{i=1}^{n}frac{1}{p^{b_i}}$, when $p geq 2$.
$begingroup$
Suppose we are given $sum_{i=1}^{n}frac{1}{a_i} geq sum_{i=1}^{n}frac{1}{b_i}$, where $a_i,b_i >0$, $forall i$. Is $sum_{i=1}^{n}frac{1}{p^{a_i}} geq sum_{i=1}^{n}frac{1}{p^{b_i}}$, where $p geq 2$?
The above seems to be intuitively true, but I haven't found a rigorous way to prove (or disprove) the above result. Any suggestions or pointers to relevant inequalities would be much appreciated.
inequality
asked Jan 31 at 5:42
BrianBrian
64
$endgroup$
add a comment |
$begingroup$
Suppose we are given $sum_{i=1}^{n}frac{1}{a_i} geq sum_{i=1}^{n}frac{1}{b_i}$, where $a_i,b_i >0$, $forall i$. Is $sum_{i=1}^{n}frac{1}{p^{a_i}} geq sum_{i=1}^{n}frac{1}{p^{b_i}}$, where $p geq 2$?
The above seems to be intuitively true, but I haven't found a rigorous way to prove (or disprove) the above result. Any suggestions or pointers to relevant inequalities would be much appreciated.
inequality
asked Jan 31 at 5:42
BrianBrian
64
$endgroup$
add a comment |
$begingroup$
Suppose we are given $sum_{i=1}^{n}frac{1}{a_i} geq sum_{i=1}^{n}frac{1}{b_i}$, where $a_i,b_i >0$, $forall i$. Is $sum_{i=1}^{n}frac{1}{p^{a_i}} geq sum_{i=1}^{n}frac{1}{p^{b_i}}$, where $p geq 2$?
The above seems to be intuitively true, but I haven't found a rigorous way to prove (or disprove) the above result. Any suggestions or pointers to relevant inequalities would be much appreciated.
inequality
asked Jan 31 at 5:42
BrianBrian
64
$endgroup$
Suppose we are given $sum_{i=1}^{n}frac{1}{a_i} geq sum_{i=1}^{n}frac{1}{b_i}$, where $a_i,b_i >0$, $forall i$. Is $sum_{i=1}^{n}frac{1}{p^{a_i}} geq sum_{i=1}^{n}frac{1}{p^{b_i}}$, where $p geq 2$?
The above seems to be intuitively true, but I haven't found a rigorous way to prove (or disprove) the above result. Any suggestions or pointers to relevant inequalities would be much appreciated.
inequality
inequality
asked Jan 31 at 5:42
BrianBrian
64
asked Jan 31 at 5:42
BrianBrian
64
asked Jan 31 at 5:42
BrianBrian
64
asked Jan 31 at 5:42
BrianBrian
64
asked Jan 31 at 5:42
BrianBrian
64
64
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, it's not necessarily true. For example, let $n = 2$, $a_1 = 0.5$, $a_2 = 100$, $b_1 = b_2 = 1$ and $p = 2$. In this case,
$$sum_{i = 1}^{n} frac{1}{a_i} = 2 + .01 = 2.01 tag{1}label{eq1}$$
$$sum_{i = 1}^{n} frac{1}{b_i} = 1 + 1 = 2 tag{2}label{eq2}$$
Thus, your first condition holds. However,
$$sum_{i = 1}^{n} frac{1}{p^{a_i}} = cfrac{1}{sqrt{2}} + cfrac{1}{2^{100}} < 1 tag{3}label{eq3}$$
but
$$sum_{i = 1}^{n} frac{1}{p^{b_i}} = cfrac{1}{2} + cfrac{1}{2} = 1 tag{4}label{eq4}$$
As such, your second condition doesn't hold. I don't believe there is any minimum value of $p$ you would allow your second condition to hold in all cases where the first condition holds, with the answer of Doyun Nam giving an example of how you can show this.
answered Jan 31 at 6:07
John OmielanJohn Omielan
4,6312215
$endgroup$
$begingroup$
Thanks. I am wondering if there is any non-trivial relation between the a_i's and the b_i's that makes this inequality true for all p greater than 2.
$endgroup$
– Brian
Jan 31 at 14:30
$begingroup$
@Brian Yes, there actually is. One simple one which is guaranteed to work is that they be positive & that each $a_i le b_i$, so $frac{1}{a_i} ge frac{1}{b_i}$. Your first inequality will then work. Also, then $p^{a_i} le p^{b_i}$, giving that $1/p^{a_i} ge 1/p^{b_i}$, so your second inequality will also work. The reason why your original statement is not always true is that you were only considering the summation. If you don't require each element of $a$ to be $le$ each $b$, you can generally (but I'm don't think always) find values of $p$ which cause your second summation to fail.
$endgroup$
– John Omielan
Jan 31 at 15:57
$begingroup$
Yes, the case where the ordering holds between the a_i's and the b_i's is fairly straightforward. I am mainly interested in knowing if the result holds for more general scenarios as well. We can restrict our attention to integer a_i's, b_i's and p, and impose upper bounds on the p.
$endgroup$
– Brian
Jan 31 at 16:55
$begingroup$
@Brian This is not my area of expertise, but I believe even with integer $a_i$, $b_i$ and $p$, including with upper bounds on $p$, the majority of situations will not hold. However, I believe some will, although it is likely difficult to quantify this, especially for larger $n$, as basically you are dealing with higher power polynomials. If you're really interested, I suggest you write another question here, stating it's an extension to this one & linking to it, where you can ask if anybody can provide any particular general conditions for your two equalities to hold.
$endgroup$
– John Omielan
Jan 31 at 19:57
$begingroup$
Okay, thanks for the suggestion.
$endgroup$
– Brian
Jan 31 at 20:23
add a comment |
$begingroup$
Your inequality isn't true in general case.
Assume $n=3$, $a_1 = a_2 = a_3 = 2$, and $b_1=1, b_2=b_3 = 4$. Then your condition for $a_i$ and $b_i$ is satisfied.
However, if your inequality is true, then
$$frac{3}{p^2} geq frac{1}{p} + frac{2}{p^4}$$
should hold for all $p geq 2$.
But it doesn't, one of the counter example is the case $p=3$.
It is another approach.
If we allow an extented real number system, let $n=2$, $a_1=a_2=2$, and $b_1=1, b_2 = infty$.
Then $frac{1}{2} + frac{1}{2} = frac{1}{1} + frac{1}{infty}$.
Thus the condition for $a_i$ and $b_i$ is satisfied.
However,
$$frac{2}{p^2} geq frac{1}{p} + frac{1}{p^infty} = frac{1}{p}$$
only holds where $2 geq p$.
answered Jan 31 at 6:15
Doyun NamDoyun Nam
66619
$endgroup$
1
$begingroup$
Nice counterexamples, thanks!
$endgroup$
– Brian
Jan 31 at 14:30
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094560%2fproving-sum-i-1n-frac1pa-i-geq-sum-i-1n-frac1pb-i%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, it's not necessarily true. For example, let $n = 2$, $a_1 = 0.5$, $a_2 = 100$, $b_1 = b_2 = 1$ and $p = 2$. In this case,
$$sum_{i = 1}^{n} frac{1}{a_i} = 2 + .01 = 2.01 tag{1}label{eq1}$$
$$sum_{i = 1}^{n} frac{1}{b_i} = 1 + 1 = 2 tag{2}label{eq2}$$
Thus, your first condition holds. However,
$$sum_{i = 1}^{n} frac{1}{p^{a_i}} = cfrac{1}{sqrt{2}} + cfrac{1}{2^{100}} < 1 tag{3}label{eq3}$$
but
$$sum_{i = 1}^{n} frac{1}{p^{b_i}} = cfrac{1}{2} + cfrac{1}{2} = 1 tag{4}label{eq4}$$
As such, your second condition doesn't hold. I don't believe there is any minimum value of $p$ you would allow your second condition to hold in all cases where the first condition holds, with the answer of Doyun Nam giving an example of how you can show this.
answered Jan 31 at 6:07
John OmielanJohn Omielan
4,6312215
$endgroup$
$begingroup$
Thanks. I am wondering if there is any non-trivial relation between the a_i's and the b_i's that makes this inequality true for all p greater than 2.
$endgroup$
– Brian
Jan 31 at 14:30
$begingroup$
@Brian Yes, there actually is. One simple one which is guaranteed to work is that they be positive & that each $a_i le b_i$, so $frac{1}{a_i} ge frac{1}{b_i}$. Your first inequality will then work. Also, then $p^{a_i} le p^{b_i}$, giving that $1/p^{a_i} ge 1/p^{b_i}$, so your second inequality will also work. The reason why your original statement is not always true is that you were only considering the summation. If you don't require each element of $a$ to be $le$ each $b$, you can generally (but I'm don't think always) find values of $p$ which cause your second summation to fail.
$endgroup$
– John Omielan
Jan 31 at 15:57
$begingroup$
Yes, the case where the ordering holds between the a_i's and the b_i's is fairly straightforward. I am mainly interested in knowing if the result holds for more general scenarios as well. We can restrict our attention to integer a_i's, b_i's and p, and impose upper bounds on the p.
$endgroup$
– Brian
Jan 31 at 16:55
$begingroup$
@Brian This is not my area of expertise, but I believe even with integer $a_i$, $b_i$ and $p$, including with upper bounds on $p$, the majority of situations will not hold. However, I believe some will, although it is likely difficult to quantify this, especially for larger $n$, as basically you are dealing with higher power polynomials. If you're really interested, I suggest you write another question here, stating it's an extension to this one & linking to it, where you can ask if anybody can provide any particular general conditions for your two equalities to hold.
$endgroup$
– John Omielan
Jan 31 at 19:57
$begingroup$
Okay, thanks for the suggestion.
$endgroup$
– Brian
Jan 31 at 20:23
add a comment |
$begingroup$
No, it's not necessarily true. For example, let $n = 2$, $a_1 = 0.5$, $a_2 = 100$, $b_1 = b_2 = 1$ and $p = 2$. In this case,
$$sum_{i = 1}^{n} frac{1}{a_i} = 2 + .01 = 2.01 tag{1}label{eq1}$$
$$sum_{i = 1}^{n} frac{1}{b_i} = 1 + 1 = 2 tag{2}label{eq2}$$
Thus, your first condition holds. However,
$$sum_{i = 1}^{n} frac{1}{p^{a_i}} = cfrac{1}{sqrt{2}} + cfrac{1}{2^{100}} < 1 tag{3}label{eq3}$$
but
$$sum_{i = 1}^{n} frac{1}{p^{b_i}} = cfrac{1}{2} + cfrac{1}{2} = 1 tag{4}label{eq4}$$
As such, your second condition doesn't hold. I don't believe there is any minimum value of $p$ you would allow your second condition to hold in all cases where the first condition holds, with the answer of Doyun Nam giving an example of how you can show this.
answered Jan 31 at 6:07
John OmielanJohn Omielan
4,6312215
$endgroup$
$begingroup$
Thanks. I am wondering if there is any non-trivial relation between the a_i's and the b_i's that makes this inequality true for all p greater than 2.
$endgroup$
– Brian
Jan 31 at 14:30
$begingroup$
@Brian Yes, there actually is. One simple one which is guaranteed to work is that they be positive & that each $a_i le b_i$, so $frac{1}{a_i} ge frac{1}{b_i}$. Your first inequality will then work. Also, then $p^{a_i} le p^{b_i}$, giving that $1/p^{a_i} ge 1/p^{b_i}$, so your second inequality will also work. The reason why your original statement is not always true is that you were only considering the summation. If you don't require each element of $a$ to be $le$ each $b$, you can generally (but I'm don't think always) find values of $p$ which cause your second summation to fail.
$endgroup$
– John Omielan
Jan 31 at 15:57
$begingroup$
Yes, the case where the ordering holds between the a_i's and the b_i's is fairly straightforward. I am mainly interested in knowing if the result holds for more general scenarios as well. We can restrict our attention to integer a_i's, b_i's and p, and impose upper bounds on the p.
$endgroup$
– Brian
Jan 31 at 16:55
$begingroup$
@Brian This is not my area of expertise, but I believe even with integer $a_i$, $b_i$ and $p$, including with upper bounds on $p$, the majority of situations will not hold. However, I believe some will, although it is likely difficult to quantify this, especially for larger $n$, as basically you are dealing with higher power polynomials. If you're really interested, I suggest you write another question here, stating it's an extension to this one & linking to it, where you can ask if anybody can provide any particular general conditions for your two equalities to hold.
$endgroup$
– John Omielan
Jan 31 at 19:57
$begingroup$
Okay, thanks for the suggestion.
$endgroup$
– Brian
Jan 31 at 20:23
add a comment |
$begingroup$
No, it's not necessarily true. For example, let $n = 2$, $a_1 = 0.5$, $a_2 = 100$, $b_1 = b_2 = 1$ and $p = 2$. In this case,
$$sum_{i = 1}^{n} frac{1}{a_i} = 2 + .01 = 2.01 tag{1}label{eq1}$$
$$sum_{i = 1}^{n} frac{1}{b_i} = 1 + 1 = 2 tag{2}label{eq2}$$
Thus, your first condition holds. However,
$$sum_{i = 1}^{n} frac{1}{p^{a_i}} = cfrac{1}{sqrt{2}} + cfrac{1}{2^{100}} < 1 tag{3}label{eq3}$$
but
$$sum_{i = 1}^{n} frac{1}{p^{b_i}} = cfrac{1}{2} + cfrac{1}{2} = 1 tag{4}label{eq4}$$
As such, your second condition doesn't hold. I don't believe there is any minimum value of $p$ you would allow your second condition to hold in all cases where the first condition holds, with the answer of Doyun Nam giving an example of how you can show this.
answered Jan 31 at 6:07
John OmielanJohn Omielan
4,6312215
$endgroup$
No, it's not necessarily true. For example, let $n = 2$, $a_1 = 0.5$, $a_2 = 100$, $b_1 = b_2 = 1$ and $p = 2$. In this case,
$$sum_{i = 1}^{n} frac{1}{a_i} = 2 + .01 = 2.01 tag{1}label{eq1}$$
$$sum_{i = 1}^{n} frac{1}{b_i} = 1 + 1 = 2 tag{2}label{eq2}$$
Thus, your first condition holds. However,
$$sum_{i = 1}^{n} frac{1}{p^{a_i}} = cfrac{1}{sqrt{2}} + cfrac{1}{2^{100}} < 1 tag{3}label{eq3}$$
but
$$sum_{i = 1}^{n} frac{1}{p^{b_i}} = cfrac{1}{2} + cfrac{1}{2} = 1 tag{4}label{eq4}$$
As such, your second condition doesn't hold. I don't believe there is any minimum value of $p$ you would allow your second condition to hold in all cases where the first condition holds, with the answer of Doyun Nam giving an example of how you can show this.
answered Jan 31 at 6:07
John OmielanJohn Omielan
4,6312215
edited Jan 31 at 6:18
answered Jan 31 at 6:07
John OmielanJohn Omielan
4,6312215
answered Jan 31 at 6:07
John OmielanJohn Omielan
4,6312215
answered Jan 31 at 6:07
John OmielanJohn Omielan
4,6312215
4,6312215
$begingroup$
Thanks. I am wondering if there is any non-trivial relation between the a_i's and the b_i's that makes this inequality true for all p greater than 2.
$endgroup$
– Brian
Jan 31 at 14:30
$begingroup$
@Brian Yes, there actually is. One simple one which is guaranteed to work is that they be positive & that each $a_i le b_i$, so $frac{1}{a_i} ge frac{1}{b_i}$. Your first inequality will then work. Also, then $p^{a_i} le p^{b_i}$, giving that $1/p^{a_i} ge 1/p^{b_i}$, so your second inequality will also work. The reason why your original statement is not always true is that you were only considering the summation. If you don't require each element of $a$ to be $le$ each $b$, you can generally (but I'm don't think always) find values of $p$ which cause your second summation to fail.
$endgroup$
– John Omielan
Jan 31 at 15:57
$begingroup$
Yes, the case where the ordering holds between the a_i's and the b_i's is fairly straightforward. I am mainly interested in knowing if the result holds for more general scenarios as well. We can restrict our attention to integer a_i's, b_i's and p, and impose upper bounds on the p.
$endgroup$
– Brian
Jan 31 at 16:55
$begingroup$
@Brian This is not my area of expertise, but I believe even with integer $a_i$, $b_i$ and $p$, including with upper bounds on $p$, the majority of situations will not hold. However, I believe some will, although it is likely difficult to quantify this, especially for larger $n$, as basically you are dealing with higher power polynomials. If you're really interested, I suggest you write another question here, stating it's an extension to this one & linking to it, where you can ask if anybody can provide any particular general conditions for your two equalities to hold.
$endgroup$
– John Omielan
Jan 31 at 19:57
$begingroup$
Okay, thanks for the suggestion.
$endgroup$
– Brian
Jan 31 at 20:23
add a comment |
$begingroup$
Thanks. I am wondering if there is any non-trivial relation between the a_i's and the b_i's that makes this inequality true for all p greater than 2.
$endgroup$
– Brian
Jan 31 at 14:30
$begingroup$
@Brian Yes, there actually is. One simple one which is guaranteed to work is that they be positive & that each $a_i le b_i$, so $frac{1}{a_i} ge frac{1}{b_i}$. Your first inequality will then work. Also, then $p^{a_i} le p^{b_i}$, giving that $1/p^{a_i} ge 1/p^{b_i}$, so your second inequality will also work. The reason why your original statement is not always true is that you were only considering the summation. If you don't require each element of $a$ to be $le$ each $b$, you can generally (but I'm don't think always) find values of $p$ which cause your second summation to fail.
$endgroup$
– John Omielan
Jan 31 at 15:57
$begingroup$
Yes, the case where the ordering holds between the a_i's and the b_i's is fairly straightforward. I am mainly interested in knowing if the result holds for more general scenarios as well. We can restrict our attention to integer a_i's, b_i's and p, and impose upper bounds on the p.
$endgroup$
– Brian
Jan 31 at 16:55
$begingroup$
@Brian This is not my area of expertise, but I believe even with integer $a_i$, $b_i$ and $p$, including with upper bounds on $p$, the majority of situations will not hold. However, I believe some will, although it is likely difficult to quantify this, especially for larger $n$, as basically you are dealing with higher power polynomials. If you're really interested, I suggest you write another question here, stating it's an extension to this one & linking to it, where you can ask if anybody can provide any particular general conditions for your two equalities to hold.
$endgroup$
– John Omielan
Jan 31 at 19:57
$begingroup$
Okay, thanks for the suggestion.
$endgroup$
– Brian
Jan 31 at 20:23
$begingroup$
Thanks. I am wondering if there is any non-trivial relation between the a_i's and the b_i's that makes this inequality true for all p greater than 2.
$endgroup$
– Brian
Jan 31 at 14:30
$begingroup$
Thanks. I am wondering if there is any non-trivial relation between the a_i's and the b_i's that makes this inequality true for all p greater than 2.
$endgroup$
– Brian
Jan 31 at 14:30
$begingroup$
@Brian Yes, there actually is. One simple one which is guaranteed to work is that they be positive & that each $a_i le b_i$, so $frac{1}{a_i} ge frac{1}{b_i}$. Your first inequality will then work. Also, then $p^{a_i} le p^{b_i}$, giving that $1/p^{a_i} ge 1/p^{b_i}$, so your second inequality will also work. The reason why your original statement is not always true is that you were only considering the summation. If you don't require each element of $a$ to be $le$ each $b$, you can generally (but I'm don't think always) find values of $p$ which cause your second summation to fail.
$endgroup$
– John Omielan
Jan 31 at 15:57
$begingroup$
@Brian Yes, there actually is. One simple one which is guaranteed to work is that they be positive & that each $a_i le b_i$, so $frac{1}{a_i} ge frac{1}{b_i}$. Your first inequality will then work. Also, then $p^{a_i} le p^{b_i}$, giving that $1/p^{a_i} ge 1/p^{b_i}$, so your second inequality will also work. The reason why your original statement is not always true is that you were only considering the summation. If you don't require each element of $a$ to be $le$ each $b$, you can generally (but I'm don't think always) find values of $p$ which cause your second summation to fail.
$endgroup$
– John Omielan
Jan 31 at 15:57
$begingroup$
Yes, the case where the ordering holds between the a_i's and the b_i's is fairly straightforward. I am mainly interested in knowing if the result holds for more general scenarios as well. We can restrict our attention to integer a_i's, b_i's and p, and impose upper bounds on the p.
$endgroup$
– Brian
Jan 31 at 16:55
$begingroup$
Yes, the case where the ordering holds between the a_i's and the b_i's is fairly straightforward. I am mainly interested in knowing if the result holds for more general scenarios as well. We can restrict our attention to integer a_i's, b_i's and p, and impose upper bounds on the p.
$endgroup$
– Brian
Jan 31 at 16:55
$begingroup$
@Brian This is not my area of expertise, but I believe even with integer $a_i$, $b_i$ and $p$, including with upper bounds on $p$, the majority of situations will not hold. However, I believe some will, although it is likely difficult to quantify this, especially for larger $n$, as basically you are dealing with higher power polynomials. If you're really interested, I suggest you write another question here, stating it's an extension to this one & linking to it, where you can ask if anybody can provide any particular general conditions for your two equalities to hold.
$endgroup$
– John Omielan
Jan 31 at 19:57
$begingroup$
@Brian This is not my area of expertise, but I believe even with integer $a_i$, $b_i$ and $p$, including with upper bounds on $p$, the majority of situations will not hold. However, I believe some will, although it is likely difficult to quantify this, especially for larger $n$, as basically you are dealing with higher power polynomials. If you're really interested, I suggest you write another question here, stating it's an extension to this one & linking to it, where you can ask if anybody can provide any particular general conditions for your two equalities to hold.
$endgroup$
– John Omielan
Jan 31 at 19:57
$begingroup$
Okay, thanks for the suggestion.
$endgroup$
– Brian
Jan 31 at 20:23
$begingroup$
Okay, thanks for the suggestion.
$endgroup$
– Brian
Jan 31 at 20:23
add a comment |
$begingroup$
Your inequality isn't true in general case.
Assume $n=3$, $a_1 = a_2 = a_3 = 2$, and $b_1=1, b_2=b_3 = 4$. Then your condition for $a_i$ and $b_i$ is satisfied.
However, if your inequality is true, then
$$frac{3}{p^2} geq frac{1}{p} + frac{2}{p^4}$$
should hold for all $p geq 2$.
But it doesn't, one of the counter example is the case $p=3$.
It is another approach.
If we allow an extented real number system, let $n=2$, $a_1=a_2=2$, and $b_1=1, b_2 = infty$.
Then $frac{1}{2} + frac{1}{2} = frac{1}{1} + frac{1}{infty}$.
Thus the condition for $a_i$ and $b_i$ is satisfied.
However,
$$frac{2}{p^2} geq frac{1}{p} + frac{1}{p^infty} = frac{1}{p}$$
only holds where $2 geq p$.
answered Jan 31 at 6:15
Doyun NamDoyun Nam
66619
$endgroup$
1
$begingroup$
Nice counterexamples, thanks!
$endgroup$
– Brian
Jan 31 at 14:30
add a comment |
$begingroup$
Your inequality isn't true in general case.
Assume $n=3$, $a_1 = a_2 = a_3 = 2$, and $b_1=1, b_2=b_3 = 4$. Then your condition for $a_i$ and $b_i$ is satisfied.
However, if your inequality is true, then
$$frac{3}{p^2} geq frac{1}{p} + frac{2}{p^4}$$
should hold for all $p geq 2$.
But it doesn't, one of the counter example is the case $p=3$.
It is another approach.
If we allow an extented real number system, let $n=2$, $a_1=a_2=2$, and $b_1=1, b_2 = infty$.
Then $frac{1}{2} + frac{1}{2} = frac{1}{1} + frac{1}{infty}$.
Thus the condition for $a_i$ and $b_i$ is satisfied.
However,
$$frac{2}{p^2} geq frac{1}{p} + frac{1}{p^infty} = frac{1}{p}$$
only holds where $2 geq p$.
answered Jan 31 at 6:15
Doyun NamDoyun Nam
66619
$endgroup$
1
$begingroup$
Nice counterexamples, thanks!
$endgroup$
– Brian
Jan 31 at 14:30
add a comment |
$begingroup$
Your inequality isn't true in general case.
Assume $n=3$, $a_1 = a_2 = a_3 = 2$, and $b_1=1, b_2=b_3 = 4$. Then your condition for $a_i$ and $b_i$ is satisfied.
However, if your inequality is true, then
$$frac{3}{p^2} geq frac{1}{p} + frac{2}{p^4}$$
should hold for all $p geq 2$.
But it doesn't, one of the counter example is the case $p=3$.
It is another approach.
If we allow an extented real number system, let $n=2$, $a_1=a_2=2$, and $b_1=1, b_2 = infty$.
Then $frac{1}{2} + frac{1}{2} = frac{1}{1} + frac{1}{infty}$.
Thus the condition for $a_i$ and $b_i$ is satisfied.
However,
$$frac{2}{p^2} geq frac{1}{p} + frac{1}{p^infty} = frac{1}{p}$$
only holds where $2 geq p$.
answered Jan 31 at 6:15
Doyun NamDoyun Nam
66619
$endgroup$
Your inequality isn't true in general case.
Assume $n=3$, $a_1 = a_2 = a_3 = 2$, and $b_1=1, b_2=b_3 = 4$. Then your condition for $a_i$ and $b_i$ is satisfied.
However, if your inequality is true, then
$$frac{3}{p^2} geq frac{1}{p} + frac{2}{p^4}$$
should hold for all $p geq 2$.
But it doesn't, one of the counter example is the case $p=3$.
It is another approach.
If we allow an extented real number system, let $n=2$, $a_1=a_2=2$, and $b_1=1, b_2 = infty$.
Then $frac{1}{2} + frac{1}{2} = frac{1}{1} + frac{1}{infty}$.
Thus the condition for $a_i$ and $b_i$ is satisfied.
However,
$$frac{2}{p^2} geq frac{1}{p} + frac{1}{p^infty} = frac{1}{p}$$
only holds where $2 geq p$.
answered Jan 31 at 6:15
Doyun NamDoyun Nam
66619
edited Jan 31 at 6:31
answered Jan 31 at 6:15
Doyun NamDoyun Nam
66619
answered Jan 31 at 6:15
Doyun NamDoyun Nam
66619
answered Jan 31 at 6:15
Doyun NamDoyun Nam
66619
66619
1
$begingroup$
Nice counterexamples, thanks!
$endgroup$
– Brian
Jan 31 at 14:30
add a comment |
1
$begingroup$
Nice counterexamples, thanks!
$endgroup$
– Brian
Jan 31 at 14:30
1
1
$begingroup$
Nice counterexamples, thanks!
$endgroup$
– Brian
Jan 31 at 14:30
$begingroup$
Nice counterexamples, thanks!
$endgroup$
– Brian
Jan 31 at 14:30
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094560%2fproving-sum-i-1n-frac1pa-i-geq-sum-i-1n-frac1pb-i%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
