O inegalitate cu radicali intr-un triunghi
Moderators: Laurian Filip, Filip Chindea, Radu Titiu, maky, Cosmin Pohoata
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
O inegalitate cu radicali intr-un triunghi
\( \triangle\ ABC\ \ \Longrightarrow\ \ \fbox{\ \sum\ \sqrt{\frac a{b+c-a}}\ \ge\ sqrt{5+\frac {2R+s-3r\sqrt 3}r}\ }\ \ge\ \sqrt{5+\frac {2R}r}\ \ge\ 3 \)
Last edited by Mateescu Constantin on Tue Aug 31, 2010 11:19 pm, edited 1 time in total.
-
Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
-
Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
DA , inegalitatea de mai sus are loc in orice triunghi !
Pentru a o arata puteti proceda astfel : Se ridica inegalitatea la patrat, se folosesc identitatile cunoscute si cateva
transformari, iar in final se ajunge la inegalitatea (destul de tare zic eu) : \( \fbox{\ AI+BI+CI\ \ge\ s+3r(2-sqrt 3)\ } \) .
Aceasta din urma am demonstrat-o aici
. Si acum cateva precizari :
\( \odot \) Folosind relatiile \( \fbox{\ \sqrt{\frac{a}{b+c-a}}=\sqrt{\frac{2R}{r}}\ \cdot\ \sin\frac{A}{2}\ } \) a.s.o. , putem transforma inegalitatea de la inceput in :
\( \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \fbox{\ \sin\ \frac A2\ +\ \sin\ \frac B2\ +\ \sin\ \frac C2\ \ge\ \sqrt{\frac{5r+2R+s-3r\sqrt 3}{2R}}\ }\ \ (\ast) \)
Desi inegalitatea\( \ (\ast) \) este valabila in orice triunghi, este surprinzator faptul ca ea este mai tare decat inegalitatea
\( \fbox{\ \sin\ \frac A2\ +\ \sin\ \frac B2\ +\ \sin\ \frac C2\ \ge\ \frac 54\ +\ \frac r{2R}\ }\ \ (\ast\ast) \) care stim ca este valabila intr-un triunghi ascutitunghic
aplicand inegalitatea Popoviciu functiei concave \( \cos \) pe \( \left(0\ ,\ \frac {\pi}2\right) \) . Mai exact, inegalitatea\( \ (\ast) \) este mai tare decat
inegalitatea\( \ (\ast\ast) \) in cazul unui triunghi ascutitunghic si, mai mult in cazul unui triunghi care satisface o relatie de tipul
\( A\ge B\ge 60^{\circ}\ge C \) . Atat pot eu demonstra . Pe un program C++ se pare ca\( \ (\ast\ast) \) ramane adevarata chiar in mai
multe triunghiuri. In acest moment, folosind inegalitatea "tare"\( \ (\ast) \), suntem in masura sa imbunatatim multe inegalitati.
\( \odot \) Folosind relatiile \( \fbox{\ AI=\sqrt{2Rr}\ \cdot\ \sqrt{\frac {b+c-a}a}\ } \) a.s.o. inegalitatea \( \fbox{\ AI+BI+CI\ \ge\ s+3r(2-\sqrt 3)\ } \) devine :
\( \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \fbox{\fbox{\ \sqrt{\frac {b+c-a}a}\ +\ \sqrt{\frac {c+a-b}b}\ +\ \sqrt{\frac {a+b-c}c}\ \ge\ \frac{s+3r\cdot (2-\sqrt 3)}{\sqrt{2Rr}}\ }} \)
\( \odot \) Cateva identitati interesante : \( \fbox{\ \begin{array}{llllll}
\diamond & \sum\ \sqrt{\frac {b+c-a}a}\ =\ \sqrt{\begin{array}\frac {s^2}{2Rr} & + & \frac r{2R} & - & 4 & + & 4\cdot\sum\ \sin\ \frac A2\end{array}} \\\\\\\\\\\\\\
\diamond & AI\ +\ BI\ +\ CI\ =\ \sqrt{\begin{array}s^2 & + & r^2 & - & 8Rr & + & 8Rr\cdot\sum\ \sin\ \frac A2\end{array}} \\\\\\\\\\\\\\
\diamond & \sum\ \tan\ \frac A4\ =\ \frac {AI+BI+CI-s}r\ \ \ \ ;\ \ \ \ \sum\ \cot\ \frac A4\ =\ \frac {AI+BI+CI+s}r \end{array}\ } \)
Pentru a o arata puteti proceda astfel : Se ridica inegalitatea la patrat, se folosesc identitatile cunoscute si cateva
transformari, iar in final se ajunge la inegalitatea (destul de tare zic eu) : \( \fbox{\ AI+BI+CI\ \ge\ s+3r(2-sqrt 3)\ } \) .
Aceasta din urma am demonstrat-o aici
. Si acum cateva precizari :\( \odot \) Folosind relatiile \( \fbox{\ \sqrt{\frac{a}{b+c-a}}=\sqrt{\frac{2R}{r}}\ \cdot\ \sin\frac{A}{2}\ } \) a.s.o. , putem transforma inegalitatea de la inceput in :
\( \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \fbox{\ \sin\ \frac A2\ +\ \sin\ \frac B2\ +\ \sin\ \frac C2\ \ge\ \sqrt{\frac{5r+2R+s-3r\sqrt 3}{2R}}\ }\ \ (\ast) \)
Desi inegalitatea\( \ (\ast) \) este valabila in orice triunghi, este surprinzator faptul ca ea este mai tare decat inegalitatea
\( \fbox{\ \sin\ \frac A2\ +\ \sin\ \frac B2\ +\ \sin\ \frac C2\ \ge\ \frac 54\ +\ \frac r{2R}\ }\ \ (\ast\ast) \) care stim ca este valabila intr-un triunghi ascutitunghic
aplicand inegalitatea Popoviciu functiei concave \( \cos \) pe \( \left(0\ ,\ \frac {\pi}2\right) \) . Mai exact, inegalitatea\( \ (\ast) \) este mai tare decat
inegalitatea\( \ (\ast\ast) \) in cazul unui triunghi ascutitunghic si, mai mult in cazul unui triunghi care satisface o relatie de tipul
\( A\ge B\ge 60^{\circ}\ge C \) . Atat pot eu demonstra . Pe un program C++ se pare ca\( \ (\ast\ast) \) ramane adevarata chiar in mai
multe triunghiuri. In acest moment, folosind inegalitatea "tare"\( \ (\ast) \), suntem in masura sa imbunatatim multe inegalitati.
\( \odot \) Folosind relatiile \( \fbox{\ AI=\sqrt{2Rr}\ \cdot\ \sqrt{\frac {b+c-a}a}\ } \) a.s.o. inegalitatea \( \fbox{\ AI+BI+CI\ \ge\ s+3r(2-\sqrt 3)\ } \) devine :
\( \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \fbox{\fbox{\ \sqrt{\frac {b+c-a}a}\ +\ \sqrt{\frac {c+a-b}b}\ +\ \sqrt{\frac {a+b-c}c}\ \ge\ \frac{s+3r\cdot (2-\sqrt 3)}{\sqrt{2Rr}}\ }} \)
\( \odot \) Cateva identitati interesante : \( \fbox{\ \begin{array}{llllll}
\diamond & \sum\ \sqrt{\frac {b+c-a}a}\ =\ \sqrt{\begin{array}\frac {s^2}{2Rr} & + & \frac r{2R} & - & 4 & + & 4\cdot\sum\ \sin\ \frac A2\end{array}} \\\\\\\\\\\\\\
\diamond & AI\ +\ BI\ +\ CI\ =\ \sqrt{\begin{array}s^2 & + & r^2 & - & 8Rr & + & 8Rr\cdot\sum\ \sin\ \frac A2\end{array}} \\\\\\\\\\\\\\
\diamond & \sum\ \tan\ \frac A4\ =\ \frac {AI+BI+CI-s}r\ \ \ \ ;\ \ \ \ \sum\ \cot\ \frac A4\ =\ \frac {AI+BI+CI+s}r \end{array}\ } \)