Daca \( ABC \) este un triunghi ascutitunghic, atunci \( \sum{\sqrt{\frac{b+c-a}{a}}}\ge 3. \)
Sorin mi-a spus ca se face cu metoda multiplicatorilor Lagrange la care am ramas "masca" si de aceea am propus-o. Eu am facut-o cu multiplicatori Lagrange. Insa sunt convins ca exista si o metoda elementara.
O inegalitate a lui S. Radulescu & I.V. Maftei.
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Virgil Nicula
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O inegalitate a lui S. Radulescu & I.V. Maftei.
Last edited by Virgil Nicula on Tue Feb 02, 2010 6:40 pm, edited 2 times in total.
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Mateescu Constantin
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Mai intai sa demonstram ca intr-un triunghi ascutitunghic \( ABC \) avem inegalitatea:
\( \overline{\underline{\left\|\ \sin\frac A2\ +\ \sin\frac B2\ +\ \sin\frac C2\ \ge\ \frac 54\ +\ \frac r{2R}\ \right\|}} \) (Tudorel Lupu, G.M. 11/2007)
Dem. Aplicam inegalitatea lui Popoviciu functiei \( \underline{\cos} \) , care pe intervalul \( \left\(0\ ,\ \frac{\pi}{2}\right\) \) este concava :
\( \cos A+\cos B+\cos C+3\cos\frac{A+B+C}{3}\ \le\ 2\cos\frac{A+B}{2}+2\cos\frac{B+C}{2}+2\cos\frac{C+A}{2} \)
\( \Longleftrightarrow\ 1+\frac rR+\frac 32\ \le\ 2\sum\ \sin\frac A2\
\Longleftrightarrow\ \overline{\underline{\left\|\ \sin\frac A2\ +\ \sin\frac B2\ +\ \sin\frac C2\ \ge\ \frac 54\ +\ \frac r{2R}\ \right\|}} \)
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Acum, revenim la inegalitatea initiala pe care o ridicam la patrat : \( \left\(\sum\ \sqrt{\frac{b+c-a}a}\right\)^2\ \ge\ 9\ \Longleftrightarrow \)
\( \sum\ \frac{b+c-a}a+2\sum\sqrt{\frac{(c+a-b)(a+b-c)}{bc}}\ \ge\ 9\ \Longleftrightarrow\ \sum\ \left\(\frac ab+\frac ba\right)-3+2\sum\sqrt{\frac{4(p-b)(p-c)}{bc}}\ \ge\ 9 \)
\( \Longleftrightarrow\ \sum\ \frac{a^2+b^2}{ab}+4\sum\ \sin\frac A2\ \ge\ 12 \) . Tinand cont de inegalitatea demonstrata mai sus e suficient sa aratam ca
\( \sum\ \frac{a^2+b^2}{ab}+4\left\(\frac 54+\frac r{2R}\right\)\ \ge\ 12\ \Longleftrightarrow\ \sum\ \frac{c^2+2ab\cos C}{ab}+\frac{2r}{R}\ \ge\ 7\ \Longleftrightarrow\ \sum\ \frac{c^2}{ab}+2\sum\ \cos C+\frac {2r}R\ \ge\ 7 \)
\( \Longleftrightarrow\ \frac{a^3+b^3+c^3}{abc}+2+\frac{2r}{R}+\frac{2r}{R}\ \ge\ 7\ \Longleftrightarrow\ \frac{2p(p^2-6Rr-3r^2)}{4Rrp}+\frac{4r}{R}\ \ge\ 5\ \Longleftrightarrow\ \frac{p^2-6Rr-3r^2}{2Rr}+\frac{4r}{R}\ \ge\ 5 \)
\( \Longleftrightarrow\ \frac{p^2-6Rr-3r^2+8r^2}{2Rr}\ \ge\ 5\ \Longleftrightarrow\ p^2+5r^2\ \ge\ 16Rr\ \ \ \) O.K.
\( \overline{\underline{\left\|\ \sin\frac A2\ +\ \sin\frac B2\ +\ \sin\frac C2\ \ge\ \frac 54\ +\ \frac r{2R}\ \right\|}} \) (Tudorel Lupu, G.M. 11/2007)
Dem. Aplicam inegalitatea lui Popoviciu functiei \( \underline{\cos} \) , care pe intervalul \( \left\(0\ ,\ \frac{\pi}{2}\right\) \) este concava :
\( \cos A+\cos B+\cos C+3\cos\frac{A+B+C}{3}\ \le\ 2\cos\frac{A+B}{2}+2\cos\frac{B+C}{2}+2\cos\frac{C+A}{2} \)
\( \Longleftrightarrow\ 1+\frac rR+\frac 32\ \le\ 2\sum\ \sin\frac A2\
\Longleftrightarrow\ \overline{\underline{\left\|\ \sin\frac A2\ +\ \sin\frac B2\ +\ \sin\frac C2\ \ge\ \frac 54\ +\ \frac r{2R}\ \right\|}} \)
===================================================================================
Acum, revenim la inegalitatea initiala pe care o ridicam la patrat : \( \left\(\sum\ \sqrt{\frac{b+c-a}a}\right\)^2\ \ge\ 9\ \Longleftrightarrow \)
\( \sum\ \frac{b+c-a}a+2\sum\sqrt{\frac{(c+a-b)(a+b-c)}{bc}}\ \ge\ 9\ \Longleftrightarrow\ \sum\ \left\(\frac ab+\frac ba\right)-3+2\sum\sqrt{\frac{4(p-b)(p-c)}{bc}}\ \ge\ 9 \)
\( \Longleftrightarrow\ \sum\ \frac{a^2+b^2}{ab}+4\sum\ \sin\frac A2\ \ge\ 12 \) . Tinand cont de inegalitatea demonstrata mai sus e suficient sa aratam ca
\( \sum\ \frac{a^2+b^2}{ab}+4\left\(\frac 54+\frac r{2R}\right\)\ \ge\ 12\ \Longleftrightarrow\ \sum\ \frac{c^2+2ab\cos C}{ab}+\frac{2r}{R}\ \ge\ 7\ \Longleftrightarrow\ \sum\ \frac{c^2}{ab}+2\sum\ \cos C+\frac {2r}R\ \ge\ 7 \)
\( \Longleftrightarrow\ \frac{a^3+b^3+c^3}{abc}+2+\frac{2r}{R}+\frac{2r}{R}\ \ge\ 7\ \Longleftrightarrow\ \frac{2p(p^2-6Rr-3r^2)}{4Rrp}+\frac{4r}{R}\ \ge\ 5\ \Longleftrightarrow\ \frac{p^2-6Rr-3r^2}{2Rr}+\frac{4r}{R}\ \ge\ 5 \)
\( \Longleftrightarrow\ \frac{p^2-6Rr-3r^2+8r^2}{2Rr}\ \ge\ 5\ \Longleftrightarrow\ p^2+5r^2\ \ge\ 16Rr\ \ \ \) O.K.
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radu tanse
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Re: O inegalitate a lui S. Radulescu & I.V. Maftei.
Domnule profesor Nicula, va sugerez sa deschideti cartea "INEGALITATI. Idei si metode" de Mihai Onucu Drambe, la pagina 232, inegalitatea 119, apoi sa deconditionati conditia triunghi ascutit unghic in substitutiile a=y+z, b=z+x, c=x+y. O sa descoperiti o demonstratie elementara!
Multa sanatate!
beat`it 
Multa sanatate!
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Virgil Nicula
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Cezar Lupu
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Re: O inegalitate a lui S. Radulescu & I.V. Maftei.
Unde esti??????radu tanse wrote:Domnule profesor Nicula, va sugerez sa deschideti cartea "INEGALITATI. Idei si metode" de Mihai Onucu Drambe, la pagina 232, inegalitatea 119, apoi sa deconditionati conditia triunghi ascutit unghic in substitutiile a=y+z, b=z+x, c=x+y. O sa descoperiti o demonstratie elementara!
Multa sanatate!
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Virgil Nicula
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