$frac6{10+x^2+y^2}+frac6{10+y^2+z^2}+frac6{10+x^2+z^2}ge frac{(sqrt6+sqrt6+sqrt6)^2}{30+2(x^2+y^2+z^2)}$
$begingroup$
How to get to that on the right side?
$$frac6{10+x^2+y^2}+frac6{10+y^2+z^2}+frac6{10+x^2+z^2}ge
frac{(sqrt6+sqrt6+sqrt6)^2}{30+2(x^2+y^2+z^2)}$$
We also have $x^2+y^2+z^2 le 12$, if it's relevant.
I tried using AM-GM and AM-HM, but I can't get to that on the right side.
algebra-precalculus
$endgroup$
add a comment |
$begingroup$
How to get to that on the right side?
$$frac6{10+x^2+y^2}+frac6{10+y^2+z^2}+frac6{10+x^2+z^2}ge
frac{(sqrt6+sqrt6+sqrt6)^2}{30+2(x^2+y^2+z^2)}$$
We also have $x^2+y^2+z^2 le 12$, if it's relevant.
I tried using AM-GM and AM-HM, but I can't get to that on the right side.
algebra-precalculus
$endgroup$
$begingroup$
Do you really mean for the three terms on the left to be identical?
$endgroup$
– Steve Kass
Dec 11 '18 at 14:22
$begingroup$
It' a cyclic sum probably.
$endgroup$
– Oldboy
Dec 11 '18 at 14:23
$begingroup$
Oh. I edited it now. Sorry.
$endgroup$
– Pero
Dec 11 '18 at 14:24
1
$begingroup$
Engel's form of Cauchy Schwarz = Titu's lemma
$endgroup$
– achille hui
Dec 11 '18 at 14:29
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Dec 11 '18 at 14:34
add a comment |
$begingroup$
How to get to that on the right side?
$$frac6{10+x^2+y^2}+frac6{10+y^2+z^2}+frac6{10+x^2+z^2}ge
frac{(sqrt6+sqrt6+sqrt6)^2}{30+2(x^2+y^2+z^2)}$$
We also have $x^2+y^2+z^2 le 12$, if it's relevant.
I tried using AM-GM and AM-HM, but I can't get to that on the right side.
algebra-precalculus
$endgroup$
How to get to that on the right side?
$$frac6{10+x^2+y^2}+frac6{10+y^2+z^2}+frac6{10+x^2+z^2}ge
frac{(sqrt6+sqrt6+sqrt6)^2}{30+2(x^2+y^2+z^2)}$$
We also have $x^2+y^2+z^2 le 12$, if it's relevant.
I tried using AM-GM and AM-HM, but I can't get to that on the right side.
algebra-precalculus
algebra-precalculus
edited Dec 11 '18 at 14:23
Pero
asked Dec 11 '18 at 14:18
PeroPero
1327
1327
$begingroup$
Do you really mean for the three terms on the left to be identical?
$endgroup$
– Steve Kass
Dec 11 '18 at 14:22
$begingroup$
It' a cyclic sum probably.
$endgroup$
– Oldboy
Dec 11 '18 at 14:23
$begingroup$
Oh. I edited it now. Sorry.
$endgroup$
– Pero
Dec 11 '18 at 14:24
1
$begingroup$
Engel's form of Cauchy Schwarz = Titu's lemma
$endgroup$
– achille hui
Dec 11 '18 at 14:29
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Dec 11 '18 at 14:34
add a comment |
$begingroup$
Do you really mean for the three terms on the left to be identical?
$endgroup$
– Steve Kass
Dec 11 '18 at 14:22
$begingroup$
It' a cyclic sum probably.
$endgroup$
– Oldboy
Dec 11 '18 at 14:23
$begingroup$
Oh. I edited it now. Sorry.
$endgroup$
– Pero
Dec 11 '18 at 14:24
1
$begingroup$
Engel's form of Cauchy Schwarz = Titu's lemma
$endgroup$
– achille hui
Dec 11 '18 at 14:29
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Dec 11 '18 at 14:34
$begingroup$
Do you really mean for the three terms on the left to be identical?
$endgroup$
– Steve Kass
Dec 11 '18 at 14:22
$begingroup$
Do you really mean for the three terms on the left to be identical?
$endgroup$
– Steve Kass
Dec 11 '18 at 14:22
$begingroup$
It' a cyclic sum probably.
$endgroup$
– Oldboy
Dec 11 '18 at 14:23
$begingroup$
It' a cyclic sum probably.
$endgroup$
– Oldboy
Dec 11 '18 at 14:23
$begingroup$
Oh. I edited it now. Sorry.
$endgroup$
– Pero
Dec 11 '18 at 14:24
$begingroup$
Oh. I edited it now. Sorry.
$endgroup$
– Pero
Dec 11 '18 at 14:24
1
1
$begingroup$
Engel's form of Cauchy Schwarz = Titu's lemma
$endgroup$
– achille hui
Dec 11 '18 at 14:29
$begingroup$
Engel's form of Cauchy Schwarz = Titu's lemma
$endgroup$
– achille hui
Dec 11 '18 at 14:29
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Dec 11 '18 at 14:34
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Dec 11 '18 at 14:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$AMge HM$
$displaystyle frac{(10+x^2+y^2)+(10+y^2+z^2)+(10+z^2+x^2)}3gefrac3{frac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}}$
$displaystylefrac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}gefrac9{30+2(x^2+y^2+z^2)}$
Multiply both sides by $6$ and you have your solution.
$[(sqrt6+sqrt6+sqrt6)^2=54]$
$endgroup$
add a comment |
$begingroup$
Note that $f(t)=6/(10+t)$ is convex in $[0,+infty)$. Hence, by Jensen's inequality,
$$f(x^2+y^2)+f(x^2+z^2)+f(z^2+y^2)geq
3fleft(frac{2x^2+2y^2+2z^2}{3}right).$$
Now
$$3fleft(frac{2x^2+2y^2+2z^2}{3}right)=
frac{3cdot 6}{10+2(x^2+y^2+z^2)/3}\=
frac{3^2cdot 6}{30+2(x^2+y^2+z^2)}$$
which is exactly your RHS.
$endgroup$
$begingroup$
...which is Jensen.
$endgroup$
– Oldboy
Dec 11 '18 at 14:26
$begingroup$
Yes, by Jensen's inequality.
$endgroup$
– Robert Z
Dec 11 '18 at 14:27
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$AMge HM$
$displaystyle frac{(10+x^2+y^2)+(10+y^2+z^2)+(10+z^2+x^2)}3gefrac3{frac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}}$
$displaystylefrac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}gefrac9{30+2(x^2+y^2+z^2)}$
Multiply both sides by $6$ and you have your solution.
$[(sqrt6+sqrt6+sqrt6)^2=54]$
$endgroup$
add a comment |
$begingroup$
$AMge HM$
$displaystyle frac{(10+x^2+y^2)+(10+y^2+z^2)+(10+z^2+x^2)}3gefrac3{frac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}}$
$displaystylefrac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}gefrac9{30+2(x^2+y^2+z^2)}$
Multiply both sides by $6$ and you have your solution.
$[(sqrt6+sqrt6+sqrt6)^2=54]$
$endgroup$
add a comment |
$begingroup$
$AMge HM$
$displaystyle frac{(10+x^2+y^2)+(10+y^2+z^2)+(10+z^2+x^2)}3gefrac3{frac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}}$
$displaystylefrac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}gefrac9{30+2(x^2+y^2+z^2)}$
Multiply both sides by $6$ and you have your solution.
$[(sqrt6+sqrt6+sqrt6)^2=54]$
$endgroup$
$AMge HM$
$displaystyle frac{(10+x^2+y^2)+(10+y^2+z^2)+(10+z^2+x^2)}3gefrac3{frac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}}$
$displaystylefrac1{10+x^2+y^2}+frac1{10+y^2+z^2}+frac1{10+z^2+x^2}gefrac9{30+2(x^2+y^2+z^2)}$
Multiply both sides by $6$ and you have your solution.
$[(sqrt6+sqrt6+sqrt6)^2=54]$
edited Dec 11 '18 at 14:54
answered Dec 11 '18 at 14:30
Shubham JohriShubham Johri
5,082717
5,082717
add a comment |
add a comment |
$begingroup$
Note that $f(t)=6/(10+t)$ is convex in $[0,+infty)$. Hence, by Jensen's inequality,
$$f(x^2+y^2)+f(x^2+z^2)+f(z^2+y^2)geq
3fleft(frac{2x^2+2y^2+2z^2}{3}right).$$
Now
$$3fleft(frac{2x^2+2y^2+2z^2}{3}right)=
frac{3cdot 6}{10+2(x^2+y^2+z^2)/3}\=
frac{3^2cdot 6}{30+2(x^2+y^2+z^2)}$$
which is exactly your RHS.
$endgroup$
$begingroup$
...which is Jensen.
$endgroup$
– Oldboy
Dec 11 '18 at 14:26
$begingroup$
Yes, by Jensen's inequality.
$endgroup$
– Robert Z
Dec 11 '18 at 14:27
add a comment |
$begingroup$
Note that $f(t)=6/(10+t)$ is convex in $[0,+infty)$. Hence, by Jensen's inequality,
$$f(x^2+y^2)+f(x^2+z^2)+f(z^2+y^2)geq
3fleft(frac{2x^2+2y^2+2z^2}{3}right).$$
Now
$$3fleft(frac{2x^2+2y^2+2z^2}{3}right)=
frac{3cdot 6}{10+2(x^2+y^2+z^2)/3}\=
frac{3^2cdot 6}{30+2(x^2+y^2+z^2)}$$
which is exactly your RHS.
$endgroup$
$begingroup$
...which is Jensen.
$endgroup$
– Oldboy
Dec 11 '18 at 14:26
$begingroup$
Yes, by Jensen's inequality.
$endgroup$
– Robert Z
Dec 11 '18 at 14:27
add a comment |
$begingroup$
Note that $f(t)=6/(10+t)$ is convex in $[0,+infty)$. Hence, by Jensen's inequality,
$$f(x^2+y^2)+f(x^2+z^2)+f(z^2+y^2)geq
3fleft(frac{2x^2+2y^2+2z^2}{3}right).$$
Now
$$3fleft(frac{2x^2+2y^2+2z^2}{3}right)=
frac{3cdot 6}{10+2(x^2+y^2+z^2)/3}\=
frac{3^2cdot 6}{30+2(x^2+y^2+z^2)}$$
which is exactly your RHS.
$endgroup$
Note that $f(t)=6/(10+t)$ is convex in $[0,+infty)$. Hence, by Jensen's inequality,
$$f(x^2+y^2)+f(x^2+z^2)+f(z^2+y^2)geq
3fleft(frac{2x^2+2y^2+2z^2}{3}right).$$
Now
$$3fleft(frac{2x^2+2y^2+2z^2}{3}right)=
frac{3cdot 6}{10+2(x^2+y^2+z^2)/3}\=
frac{3^2cdot 6}{30+2(x^2+y^2+z^2)}$$
which is exactly your RHS.
edited Dec 11 '18 at 14:32
answered Dec 11 '18 at 14:25
Robert ZRobert Z
96.1k1065136
96.1k1065136
$begingroup$
...which is Jensen.
$endgroup$
– Oldboy
Dec 11 '18 at 14:26
$begingroup$
Yes, by Jensen's inequality.
$endgroup$
– Robert Z
Dec 11 '18 at 14:27
add a comment |
$begingroup$
...which is Jensen.
$endgroup$
– Oldboy
Dec 11 '18 at 14:26
$begingroup$
Yes, by Jensen's inequality.
$endgroup$
– Robert Z
Dec 11 '18 at 14:27
$begingroup$
...which is Jensen.
$endgroup$
– Oldboy
Dec 11 '18 at 14:26
$begingroup$
...which is Jensen.
$endgroup$
– Oldboy
Dec 11 '18 at 14:26
$begingroup$
Yes, by Jensen's inequality.
$endgroup$
– Robert Z
Dec 11 '18 at 14:27
$begingroup$
Yes, by Jensen's inequality.
$endgroup$
– Robert Z
Dec 11 '18 at 14:27
add a comment |
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$begingroup$
Do you really mean for the three terms on the left to be identical?
$endgroup$
– Steve Kass
Dec 11 '18 at 14:22
$begingroup$
It' a cyclic sum probably.
$endgroup$
– Oldboy
Dec 11 '18 at 14:23
$begingroup$
Oh. I edited it now. Sorry.
$endgroup$
– Pero
Dec 11 '18 at 14:24
1
$begingroup$
Engel's form of Cauchy Schwarz = Titu's lemma
$endgroup$
– achille hui
Dec 11 '18 at 14:29
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Dec 11 '18 at 14:34