Inegalitate in triunghi

[opincariumihai]

Aratati ca intre elementele unui triungi \( ABC \) are loc inegalitatea : \( \sum \frac{a}{AI^2} \geq \frac{p}{2r^2}. \)

Mihai Opincariu

[Marius Mainea]

\( LHS=\frac{1}{4R}\sum \frac{\sin A}{\sin^2\frac{B}{2}\cdot\sin^2\frac{C}{2}}\ge\frac{1}{4R}\frac{\prod\cos\frac{A}{2}}{\prod\sin^2\frac{A}{2}}=RHS \)

\( \Longleftrightarrow \) \( \sum \sin A\ge\sum \sin 2A \) \( \Longleftrightarrow \) \( \prod \sin \frac{A}{2}\le \frac{1}{8} \)

[opincariumihai]

\( \sum \frac{a}{AI^2}=\frac{p(R-r)}{r^2R}\geq \frac{p}{2r^2}. \)

[Virgil Nicula]

Inegalitatea este adevarata in orice triunghi, nu neaparat ascutitunghic !

Mi-ar fi placut sa o demonstrezi fara a folosi identitatea remarcabila \( \underline{\overline{\left\|\ \sum a^2(p-b)(p-c)=4p^2r(R-r)\ \right\|}} \) .

De exemplu, \( IA^2=\frac {bc(p-a)}{p} \) si celelalte \( \Longrightarrow\sum\frac {a}{IA^2}=\sum\frac {ap}{bc(p-a)}=\frac {p}{abc}\sum\frac {a^2}{p-a}=\frac {1}{4Rr}\sum\frac {a^2}{p-a} \).

Insa \( \sum\frac {a^2}{p-a}\stackrel{C.B.S.}{\ \ \ge\ \ }\frac {\left(\sum a\right)^2}{\sum (p-a)}=4p \).

Asadar \( \sum\frac {a}{IA^2}\ge \frac {p}{Rr} \), insa mai slaba decat cea propusa de tine deoarece \( \frac {p}{2r^2}\ge\frac {p}{Rr} \) . Cred ca m-ai inteles ...

O consecinta interesanta a acestei frumoase inegalitati este :

Sa se arate ca in orice triunghi \( ABC \) exista relatia \( \underline{\overline{\left\|\ \max\ \{\ \frac {a}{p-a}\ ,\ \frac {b}{p-b}\ ,\ \frac {c}{p-c}\ \}\ \ge \ \frac Rr\ \right\|}} \) .

O indicatie "tare" ar fi relatia \( \frac {a\cdot\frac {a}{p-a}+b\cdot\frac {b}{p-b}+c\cdot\frac {c}{p-c}}{a+b+c}\ \ge\ \frac {R}{r} \) care este chiar inegalitatea propusa de tine ...