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: $$(a^2+b^2)2ab le frac(a+b)^44$$
$$fraca^2+b^2a+b le frac(a+b)^38able frac(a+b)^3c8$$
$$sum sqrt fraca^2+b^2a+b le sum fraca+b8.2sqrt(a+b)2cle sum frac(a+b)^28+sum fracab2 le frac(a+b+c)^23$$







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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.
    – 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
    – Tsukuyomi
    Aug 18 at 1:17














up vote
-4
down vote

favorite
2












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: $$(a^2+b^2)2ab le frac(a+b)^44$$
$$fraca^2+b^2a+b le frac(a+b)^38able frac(a+b)^3c8$$
$$sum sqrt fraca^2+b^2a+b le sum fraca+b8.2sqrt(a+b)2cle sum frac(a+b)^28+sum fracab2 le frac(a+b+c)^23$$







share|cite|improve this question






















  • To get faster help, and rather than just demanding solutions, it is recommended to show what you worked out so far.
    – 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
    – Tsukuyomi
    Aug 18 at 1:17












up vote
-4
down vote

favorite
2









up vote
-4
down vote

favorite
2






2





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: $$(a^2+b^2)2ab le frac(a+b)^44$$
$$fraca^2+b^2a+b le frac(a+b)^38able frac(a+b)^3c8$$
$$sum sqrt fraca^2+b^2a+b le sum fraca+b8.2sqrt(a+b)2cle sum frac(a+b)^28+sum fracab2 le frac(a+b+c)^23$$







share|cite|improve this question














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: $$(a^2+b^2)2ab le frac(a+b)^44$$
$$fraca^2+b^2a+b le frac(a+b)^38able frac(a+b)^3c8$$
$$sum sqrt fraca^2+b^2a+b le sum fraca+b8.2sqrt(a+b)2cle sum frac(a+b)^28+sum fracab2 le frac(a+b+c)^23$$









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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.
    – 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
    – 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.
    – 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
    – 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.
– 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.
– 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
– 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
– Tsukuyomi
Aug 18 at 1:17










1 Answer
1






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!






share|cite|improve this answer






















  • It's very kind of you if you help me solve 2,3
    – 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.
    – Michael Rozenberg
    Aug 18 at 2:29










  • Thank you, I can see it, help me solve 2,3
    – Tsukuyomi
    Aug 18 at 2:30










  • Open another topic or wait that someone will help you.
    – Michael Rozenberg
    Aug 18 at 2:33










  • Why you don't help :(
    – Tsukuyomi
    Aug 18 at 2:34










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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!






share|cite|improve this answer






















  • It's very kind of you if you help me solve 2,3
    – 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.
    – Michael Rozenberg
    Aug 18 at 2:29










  • Thank you, I can see it, help me solve 2,3
    – Tsukuyomi
    Aug 18 at 2:30










  • Open another topic or wait that someone will help you.
    – Michael Rozenberg
    Aug 18 at 2:33










  • Why you don't help :(
    – Tsukuyomi
    Aug 18 at 2:34














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!






share|cite|improve this answer






















  • It's very kind of you if you help me solve 2,3
    – 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.
    – Michael Rozenberg
    Aug 18 at 2:29










  • Thank you, I can see it, help me solve 2,3
    – Tsukuyomi
    Aug 18 at 2:30










  • Open another topic or wait that someone will help you.
    – Michael Rozenberg
    Aug 18 at 2:33










  • Why you don't help :(
    – Tsukuyomi
    Aug 18 at 2:34












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!






share|cite|improve this answer














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!







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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
    – 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.
    – Michael Rozenberg
    Aug 18 at 2:29










  • Thank you, I can see it, help me solve 2,3
    – Tsukuyomi
    Aug 18 at 2:30










  • Open another topic or wait that someone will help you.
    – Michael Rozenberg
    Aug 18 at 2:33










  • Why you don't help :(
    – Tsukuyomi
    Aug 18 at 2:34
















  • It's very kind of you if you help me solve 2,3
    – 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.
    – Michael Rozenberg
    Aug 18 at 2:29










  • Thank you, I can see it, help me solve 2,3
    – Tsukuyomi
    Aug 18 at 2:30










  • Open another topic or wait that someone will help you.
    – Michael Rozenberg
    Aug 18 at 2:33










  • Why you don't help :(
    – Tsukuyomi
    Aug 18 at 2:34















It's very kind of you if you help me solve 2,3
– Tsukuyomi
Aug 18 at 2:26





It's very kind of you if you help me solve 2,3
– 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.
– 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.
– Michael Rozenberg
Aug 18 at 2:29












Thank you, I can see it, help me solve 2,3
– Tsukuyomi
Aug 18 at 2:30




Thank you, I can see it, help me solve 2,3
– Tsukuyomi
Aug 18 at 2:30












Open another topic or wait that someone will help you.
– Michael Rozenberg
Aug 18 at 2:33




Open another topic or wait that someone will help you.
– Michael Rozenberg
Aug 18 at 2:33












Why you don't help :(
– Tsukuyomi
Aug 18 at 2:34




Why you don't help :(
– Tsukuyomi
Aug 18 at 2:34












 

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