Inequality by Cauchy-Schwarz and uvw
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Problem 2: Let $a,b,c>0$, $abcge 1$. Prove that
$sum sqrt fraca^2+b^2a+b le frac(a+b+c)^23$
My solution for problem 2: $$(a^2+b^2)2ab le frac(a+b)^44$$
$$fraca^2+b^2a+b le frac(a+b)^38able frac(a+b)^3c8$$
$$sum sqrt fraca^2+b^2a+b le sum fraca+b8.2sqrt(a+b)2cle sum frac(a+b)^28+sum fracab2 le frac(a+b+c)^23$$
inequality radicals
add a comment |Â
up vote
-4
down vote
favorite
Problem 2: Let $a,b,c>0$, $abcge 1$. Prove that
$sum sqrt fraca^2+b^2a+b le frac(a+b+c)^23$
My solution for problem 2: $$(a^2+b^2)2ab le frac(a+b)^44$$
$$fraca^2+b^2a+b le frac(a+b)^38able frac(a+b)^3c8$$
$$sum sqrt fraca^2+b^2a+b le sum fraca+b8.2sqrt(a+b)2cle sum frac(a+b)^28+sum fracab2 le frac(a+b+c)^23$$
inequality radicals
To get faster help, and rather than just demanding solutions, it is recommended to show what you worked out so far.
â Ahmad Bazzi
Aug 18 at 1:14
I know the fist use EV and Schur for third, but it is very long, and I don't succeed
â Tsukuyomi
Aug 18 at 1:17
add a comment |Â
up vote
-4
down vote
favorite
up vote
-4
down vote
favorite
Problem 2: Let $a,b,c>0$, $abcge 1$. Prove that
$sum sqrt fraca^2+b^2a+b le frac(a+b+c)^23$
My solution for problem 2: $$(a^2+b^2)2ab le frac(a+b)^44$$
$$fraca^2+b^2a+b le frac(a+b)^38able frac(a+b)^3c8$$
$$sum sqrt fraca^2+b^2a+b le sum fraca+b8.2sqrt(a+b)2cle sum frac(a+b)^28+sum fracab2 le frac(a+b+c)^23$$
inequality radicals
Problem 2: Let $a,b,c>0$, $abcge 1$. Prove that
$sum sqrt fraca^2+b^2a+b le frac(a+b+c)^23$
My solution for problem 2: $$(a^2+b^2)2ab le frac(a+b)^44$$
$$fraca^2+b^2a+b le frac(a+b)^38able frac(a+b)^3c8$$
$$sum sqrt fraca^2+b^2a+b le sum fraca+b8.2sqrt(a+b)2cle sum frac(a+b)^28+sum fracab2 le frac(a+b+c)^23$$
inequality radicals
edited Aug 18 at 8:04
asked Aug 18 at 1:09
Tsukuyomi
275
275
To get faster help, and rather than just demanding solutions, it is recommended to show what you worked out so far.
â Ahmad Bazzi
Aug 18 at 1:14
I know the fist use EV and Schur for third, but it is very long, and I don't succeed
â Tsukuyomi
Aug 18 at 1:17
add a comment |Â
To get faster help, and rather than just demanding solutions, it is recommended to show what you worked out so far.
â Ahmad Bazzi
Aug 18 at 1:14
I know the fist use EV and Schur for third, but it is very long, and I don't succeed
â Tsukuyomi
Aug 18 at 1:17
To get faster help, and rather than just demanding solutions, it is recommended to show what you worked out so far.
â Ahmad Bazzi
Aug 18 at 1:14
To get faster help, and rather than just demanding solutions, it is recommended to show what you worked out so far.
â Ahmad Bazzi
Aug 18 at 1:14
I know the fist use EV and Schur for third, but it is very long, and I don't succeed
â Tsukuyomi
Aug 18 at 1:17
I know the fist use EV and Schur for third, but it is very long, and I don't succeed
â Tsukuyomi
Aug 18 at 1:17
add a comment |Â
1 Answer
1
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oldest
votes
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1
down vote
For the first problem the Vasc's EV Method indeed helps!
Let $a^2+b^2+c^2+d^2=constant.$
Thus, by Corollary 1.7(b) here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf
it's enough to prove our inequality for an equality case of three variables.
Let $b=c=a$.
Hence, $d=4-3a$, which gives $0<a<frac43$ and we need to prove that
$$4096geq243a^3(4-3a)(3a^2+(4-3a)^2)^2$$ or $f(a)geq0,$ where
$$f(a)=8ln2-5ln3-3lna-ln(4-3a)-2ln(3a^2-6a+4).$$
Now, $$f'(a)=frac24(a-1)^2(3a-2)a(4-3a)(3a^2-6a+4),$$ which gives $a_min=frac23$ and since $fleft(frac23right)=0,$ we are done!
It's very kind of you if you help me solve 2,3
â Tsukuyomi
Aug 18 at 2:26
@Tsukuyomi I checked and I see it very well. If so try in the google: Vasile Cirtoaje EV method.
â Michael Rozenberg
Aug 18 at 2:29
Thank you, I can see it, help me solve 2,3
â Tsukuyomi
Aug 18 at 2:30
Open another topic or wait that someone will help you.
â Michael Rozenberg
Aug 18 at 2:33
Why you don't help :(
â Tsukuyomi
Aug 18 at 2:34
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
For the first problem the Vasc's EV Method indeed helps!
Let $a^2+b^2+c^2+d^2=constant.$
Thus, by Corollary 1.7(b) here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf
it's enough to prove our inequality for an equality case of three variables.
Let $b=c=a$.
Hence, $d=4-3a$, which gives $0<a<frac43$ and we need to prove that
$$4096geq243a^3(4-3a)(3a^2+(4-3a)^2)^2$$ or $f(a)geq0,$ where
$$f(a)=8ln2-5ln3-3lna-ln(4-3a)-2ln(3a^2-6a+4).$$
Now, $$f'(a)=frac24(a-1)^2(3a-2)a(4-3a)(3a^2-6a+4),$$ which gives $a_min=frac23$ and since $fleft(frac23right)=0,$ we are done!
It's very kind of you if you help me solve 2,3
â Tsukuyomi
Aug 18 at 2:26
@Tsukuyomi I checked and I see it very well. If so try in the google: Vasile Cirtoaje EV method.
â Michael Rozenberg
Aug 18 at 2:29
Thank you, I can see it, help me solve 2,3
â Tsukuyomi
Aug 18 at 2:30
Open another topic or wait that someone will help you.
â Michael Rozenberg
Aug 18 at 2:33
Why you don't help :(
â Tsukuyomi
Aug 18 at 2:34
 |Â
show 4 more comments
up vote
1
down vote
For the first problem the Vasc's EV Method indeed helps!
Let $a^2+b^2+c^2+d^2=constant.$
Thus, by Corollary 1.7(b) here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf
it's enough to prove our inequality for an equality case of three variables.
Let $b=c=a$.
Hence, $d=4-3a$, which gives $0<a<frac43$ and we need to prove that
$$4096geq243a^3(4-3a)(3a^2+(4-3a)^2)^2$$ or $f(a)geq0,$ where
$$f(a)=8ln2-5ln3-3lna-ln(4-3a)-2ln(3a^2-6a+4).$$
Now, $$f'(a)=frac24(a-1)^2(3a-2)a(4-3a)(3a^2-6a+4),$$ which gives $a_min=frac23$ and since $fleft(frac23right)=0,$ we are done!
It's very kind of you if you help me solve 2,3
â Tsukuyomi
Aug 18 at 2:26
@Tsukuyomi I checked and I see it very well. If so try in the google: Vasile Cirtoaje EV method.
â Michael Rozenberg
Aug 18 at 2:29
Thank you, I can see it, help me solve 2,3
â Tsukuyomi
Aug 18 at 2:30
Open another topic or wait that someone will help you.
â Michael Rozenberg
Aug 18 at 2:33
Why you don't help :(
â Tsukuyomi
Aug 18 at 2:34
 |Â
show 4 more comments
up vote
1
down vote
up vote
1
down vote
For the first problem the Vasc's EV Method indeed helps!
Let $a^2+b^2+c^2+d^2=constant.$
Thus, by Corollary 1.7(b) here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf
it's enough to prove our inequality for an equality case of three variables.
Let $b=c=a$.
Hence, $d=4-3a$, which gives $0<a<frac43$ and we need to prove that
$$4096geq243a^3(4-3a)(3a^2+(4-3a)^2)^2$$ or $f(a)geq0,$ where
$$f(a)=8ln2-5ln3-3lna-ln(4-3a)-2ln(3a^2-6a+4).$$
Now, $$f'(a)=frac24(a-1)^2(3a-2)a(4-3a)(3a^2-6a+4),$$ which gives $a_min=frac23$ and since $fleft(frac23right)=0,$ we are done!
For the first problem the Vasc's EV Method indeed helps!
Let $a^2+b^2+c^2+d^2=constant.$
Thus, by Corollary 1.7(b) here
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf
it's enough to prove our inequality for an equality case of three variables.
Let $b=c=a$.
Hence, $d=4-3a$, which gives $0<a<frac43$ and we need to prove that
$$4096geq243a^3(4-3a)(3a^2+(4-3a)^2)^2$$ or $f(a)geq0,$ where
$$f(a)=8ln2-5ln3-3lna-ln(4-3a)-2ln(3a^2-6a+4).$$
Now, $$f'(a)=frac24(a-1)^2(3a-2)a(4-3a)(3a^2-6a+4),$$ which gives $a_min=frac23$ and since $fleft(frac23right)=0,$ we are done!
edited Aug 18 at 7:18
answered Aug 18 at 2:24
Michael Rozenberg
88.5k1579179
88.5k1579179
It's very kind of you if you help me solve 2,3
â Tsukuyomi
Aug 18 at 2:26
@Tsukuyomi I checked and I see it very well. If so try in the google: Vasile Cirtoaje EV method.
â Michael Rozenberg
Aug 18 at 2:29
Thank you, I can see it, help me solve 2,3
â Tsukuyomi
Aug 18 at 2:30
Open another topic or wait that someone will help you.
â Michael Rozenberg
Aug 18 at 2:33
Why you don't help :(
â Tsukuyomi
Aug 18 at 2:34
 |Â
show 4 more comments
It's very kind of you if you help me solve 2,3
â Tsukuyomi
Aug 18 at 2:26
@Tsukuyomi I checked and I see it very well. If so try in the google: Vasile Cirtoaje EV method.
â Michael Rozenberg
Aug 18 at 2:29
Thank you, I can see it, help me solve 2,3
â Tsukuyomi
Aug 18 at 2:30
Open another topic or wait that someone will help you.
â Michael Rozenberg
Aug 18 at 2:33
Why you don't help :(
â Tsukuyomi
Aug 18 at 2:34
It's very kind of you if you help me solve 2,3
â Tsukuyomi
Aug 18 at 2:26
It's very kind of you if you help me solve 2,3
â Tsukuyomi
Aug 18 at 2:26
@Tsukuyomi I checked and I see it very well. If so try in the google: Vasile Cirtoaje EV method.
â Michael Rozenberg
Aug 18 at 2:29
@Tsukuyomi I checked and I see it very well. If so try in the google: Vasile Cirtoaje EV method.
â Michael Rozenberg
Aug 18 at 2:29
Thank you, I can see it, help me solve 2,3
â Tsukuyomi
Aug 18 at 2:30
Thank you, I can see it, help me solve 2,3
â Tsukuyomi
Aug 18 at 2:30
Open another topic or wait that someone will help you.
â Michael Rozenberg
Aug 18 at 2:33
Open another topic or wait that someone will help you.
â Michael Rozenberg
Aug 18 at 2:33
Why you don't help :(
â Tsukuyomi
Aug 18 at 2:34
Why you don't help :(
â Tsukuyomi
Aug 18 at 2:34
 |Â
show 4 more comments
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To get faster help, and rather than just demanding solutions, it is recommended to show what you worked out so far.
â Ahmad Bazzi
Aug 18 at 1:14
I know the fist use EV and Schur for third, but it is very long, and I don't succeed
â Tsukuyomi
Aug 18 at 1:17