Inequality proof $frac{x}{y}+2frac{y}{z}+3frac{z}{x} geq 3sqrt[3]{6} $ [closed]
$begingroup$
I have found numerically that for $x>0,y>0,z>0$ we have
$$
frac{x}{y}+2frac{y}{z}+3frac{z}{x} geq 3sqrt[3]{6}
$$
but I don't know how to prove this. Do you have any ideas?
inequality
edited Dec 1 '18 at 16:05
Martin Sleziak
44.7k8117272
asked Dec 1 '18 at 12:16
avan1235avan1235
1926
$endgroup$
closed as off-topic by Saad, Martin Sleziak, Did, user10354138, KReiser Dec 2 '18 at 6:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Martin Sleziak, Did, user10354138, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have found numerically that for $x>0,y>0,z>0$ we have
$$
frac{x}{y}+2frac{y}{z}+3frac{z}{x} geq 3sqrt[3]{6}
$$
but I don't know how to prove this. Do you have any ideas?
inequality
edited Dec 1 '18 at 16:05
Martin Sleziak
44.7k8117272
asked Dec 1 '18 at 12:16
avan1235avan1235
1926
$endgroup$
closed as off-topic by Saad, Martin Sleziak, Did, user10354138, KReiser Dec 2 '18 at 6:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Martin Sleziak, Did, user10354138, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
5
$begingroup$
Immediate from the AM-GM inequality.
$endgroup$
– user10354138
Dec 1 '18 at 12:19
$begingroup$
I am curious, how did you find this, "numerically"? Please be specific.
$endgroup$
– Did
Dec 1 '18 at 17:13
add a comment |
$begingroup$
I have found numerically that for $x>0,y>0,z>0$ we have
$$
frac{x}{y}+2frac{y}{z}+3frac{z}{x} geq 3sqrt[3]{6}
$$
but I don't know how to prove this. Do you have any ideas?
inequality
edited Dec 1 '18 at 16:05
Martin Sleziak
44.7k8117272
asked Dec 1 '18 at 12:16
avan1235avan1235
1926
$endgroup$
I have found numerically that for $x>0,y>0,z>0$ we have
$$
frac{x}{y}+2frac{y}{z}+3frac{z}{x} geq 3sqrt[3]{6}
$$
but I don't know how to prove this. Do you have any ideas?
inequality
inequality
edited Dec 1 '18 at 16:05
Martin Sleziak
44.7k8117272
asked Dec 1 '18 at 12:16
avan1235avan1235
1926
edited Dec 1 '18 at 16:05
Martin Sleziak
44.7k8117272
asked Dec 1 '18 at 12:16
avan1235avan1235
1926
edited Dec 1 '18 at 16:05
Martin Sleziak
44.7k8117272
edited Dec 1 '18 at 16:05
Martin Sleziak
44.7k8117272
edited Dec 1 '18 at 16:05
Martin Sleziak
44.7k8117272
44.7k8117272
asked Dec 1 '18 at 12:16
avan1235avan1235
1926
asked Dec 1 '18 at 12:16
avan1235avan1235
1926
asked Dec 1 '18 at 12:16
avan1235avan1235
1926
1926
closed as off-topic by Saad, Martin Sleziak, Did, user10354138, KReiser Dec 2 '18 at 6:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Martin Sleziak, Did, user10354138, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Martin Sleziak, Did, user10354138, KReiser Dec 2 '18 at 6:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Martin Sleziak, Did, user10354138, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
5
$begingroup$
Immediate from the AM-GM inequality.
$endgroup$
– user10354138
Dec 1 '18 at 12:19
$begingroup$
I am curious, how did you find this, "numerically"? Please be specific.
$endgroup$
– Did
Dec 1 '18 at 17:13
add a comment |
5
$begingroup$
Immediate from the AM-GM inequality.
$endgroup$
– user10354138
Dec 1 '18 at 12:19
$begingroup$
I am curious, how did you find this, "numerically"? Please be specific.
$endgroup$
– Did
Dec 1 '18 at 17:13
5
5
$begingroup$
Immediate from the AM-GM inequality.
$endgroup$
– user10354138
Dec 1 '18 at 12:19
$begingroup$
Immediate from the AM-GM inequality.
$endgroup$
– user10354138
Dec 1 '18 at 12:19
$begingroup$
I am curious, how did you find this, "numerically"? Please be specific.
$endgroup$
– Did
Dec 1 '18 at 17:13
$begingroup$
I am curious, how did you find this, "numerically"? Please be specific.
$endgroup$
– Did
Dec 1 '18 at 17:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's quite interesting you found it numerically. How did you do it?
However you can prove your observation using the lemma $$a+b+c ge 3 sqrt[3]{abc}$$
For $a=dfrac{x}{y}$ , $b=dfrac{2y}{z}$ and $c=dfrac{3z}{x}$ the prove follows.
answered Dec 1 '18 at 12:43
Atiq RahmanAtiq Rahman
1143
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's quite interesting you found it numerically. How did you do it?
However you can prove your observation using the lemma $$a+b+c ge 3 sqrt[3]{abc}$$
For $a=dfrac{x}{y}$ , $b=dfrac{2y}{z}$ and $c=dfrac{3z}{x}$ the prove follows.
answered Dec 1 '18 at 12:43
Atiq RahmanAtiq Rahman
1143
$endgroup$
add a comment |
$begingroup$
It's quite interesting you found it numerically. How did you do it?
However you can prove your observation using the lemma $$a+b+c ge 3 sqrt[3]{abc}$$
For $a=dfrac{x}{y}$ , $b=dfrac{2y}{z}$ and $c=dfrac{3z}{x}$ the prove follows.
answered Dec 1 '18 at 12:43
Atiq RahmanAtiq Rahman
1143
$endgroup$
add a comment |
$begingroup$
It's quite interesting you found it numerically. How did you do it?
However you can prove your observation using the lemma $$a+b+c ge 3 sqrt[3]{abc}$$
For $a=dfrac{x}{y}$ , $b=dfrac{2y}{z}$ and $c=dfrac{3z}{x}$ the prove follows.
answered Dec 1 '18 at 12:43
Atiq RahmanAtiq Rahman
1143
$endgroup$
It's quite interesting you found it numerically. How did you do it?
However you can prove your observation using the lemma $$a+b+c ge 3 sqrt[3]{abc}$$
For $a=dfrac{x}{y}$ , $b=dfrac{2y}{z}$ and $c=dfrac{3z}{x}$ the prove follows.
answered Dec 1 '18 at 12:43
Atiq RahmanAtiq Rahman
1143
answered Dec 1 '18 at 12:43
Atiq RahmanAtiq Rahman
1143
answered Dec 1 '18 at 12:43
Atiq RahmanAtiq Rahman
1143
answered Dec 1 '18 at 12:43
Atiq RahmanAtiq Rahman
1143
1143
add a comment |
add a comment |
5
$begingroup$
Immediate from the AM-GM inequality.
$endgroup$
– user10354138
Dec 1 '18 at 12:19
$begingroup$
I am curious, how did you find this, "numerically"? Please be specific.
$endgroup$
– Did
Dec 1 '18 at 17:13