A geometric inequality to triangle $ABC$
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If $a, b, c$ be the sides of a triangle $ABC$, I have to prove that
$sumlimits_cycl^ (frac ab+c)^2 + frac 3 A^2abcs ge frac 98$
where $A,s$ the area and the semiperemeter. I tried to use the Ravi's substitution, but the inequality was then quite hard to solve. It might be a differente method. Thank you
inequality contest-math
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up vote
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If $a, b, c$ be the sides of a triangle $ABC$, I have to prove that
$sumlimits_cycl^ (frac ab+c)^2 + frac 3 A^2abcs ge frac 98$
where $A,s$ the area and the semiperemeter. I tried to use the Ravi's substitution, but the inequality was then quite hard to solve. It might be a differente method. Thank you
inequality contest-math
1
Use the $uvw$ method.
– Michael Rozenberg
Jul 19 at 14:03
3
Isn't this inequality equivalent to the one you asked in another question?
– achille hui
Jul 19 at 14:11
1
Yes of course, but he wants another method. See please better a starting post.
– Michael Rozenberg
Jul 19 at 14:16
1
@Steven I found a proof by SOS, but it's very ugly.
– Michael Rozenberg
Jul 19 at 15:57
Thank you @MichaelRozenberg, you are very strong
– Steven
Jul 19 at 20:44
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $a, b, c$ be the sides of a triangle $ABC$, I have to prove that
$sumlimits_cycl^ (frac ab+c)^2 + frac 3 A^2abcs ge frac 98$
where $A,s$ the area and the semiperemeter. I tried to use the Ravi's substitution, but the inequality was then quite hard to solve. It might be a differente method. Thank you
inequality contest-math
If $a, b, c$ be the sides of a triangle $ABC$, I have to prove that
$sumlimits_cycl^ (frac ab+c)^2 + frac 3 A^2abcs ge frac 98$
where $A,s$ the area and the semiperemeter. I tried to use the Ravi's substitution, but the inequality was then quite hard to solve. It might be a differente method. Thank you
inequality contest-math
edited 19 hours ago


amWhy
189k25219431
189k25219431
asked Jul 19 at 13:33
Steven
41019
41019
1
Use the $uvw$ method.
– Michael Rozenberg
Jul 19 at 14:03
3
Isn't this inequality equivalent to the one you asked in another question?
– achille hui
Jul 19 at 14:11
1
Yes of course, but he wants another method. See please better a starting post.
– Michael Rozenberg
Jul 19 at 14:16
1
@Steven I found a proof by SOS, but it's very ugly.
– Michael Rozenberg
Jul 19 at 15:57
Thank you @MichaelRozenberg, you are very strong
– Steven
Jul 19 at 20:44
add a comment |Â
1
Use the $uvw$ method.
– Michael Rozenberg
Jul 19 at 14:03
3
Isn't this inequality equivalent to the one you asked in another question?
– achille hui
Jul 19 at 14:11
1
Yes of course, but he wants another method. See please better a starting post.
– Michael Rozenberg
Jul 19 at 14:16
1
@Steven I found a proof by SOS, but it's very ugly.
– Michael Rozenberg
Jul 19 at 15:57
Thank you @MichaelRozenberg, you are very strong
– Steven
Jul 19 at 20:44
1
1
Use the $uvw$ method.
– Michael Rozenberg
Jul 19 at 14:03
Use the $uvw$ method.
– Michael Rozenberg
Jul 19 at 14:03
3
3
Isn't this inequality equivalent to the one you asked in another question?
– achille hui
Jul 19 at 14:11
Isn't this inequality equivalent to the one you asked in another question?
– achille hui
Jul 19 at 14:11
1
1
Yes of course, but he wants another method. See please better a starting post.
– Michael Rozenberg
Jul 19 at 14:16
Yes of course, but he wants another method. See please better a starting post.
– Michael Rozenberg
Jul 19 at 14:16
1
1
@Steven I found a proof by SOS, but it's very ugly.
– Michael Rozenberg
Jul 19 at 15:57
@Steven I found a proof by SOS, but it's very ugly.
– Michael Rozenberg
Jul 19 at 15:57
Thank you @MichaelRozenberg, you are very strong
– Steven
Jul 19 at 20:44
Thank you @MichaelRozenberg, you are very strong
– Steven
Jul 19 at 20:44
add a comment |Â
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1
Use the $uvw$ method.
– Michael Rozenberg
Jul 19 at 14:03
3
Isn't this inequality equivalent to the one you asked in another question?
– achille hui
Jul 19 at 14:11
1
Yes of course, but he wants another method. See please better a starting post.
– Michael Rozenberg
Jul 19 at 14:16
1
@Steven I found a proof by SOS, but it's very ugly.
– Michael Rozenberg
Jul 19 at 15:57
Thank you @MichaelRozenberg, you are very strong
– Steven
Jul 19 at 20:44