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Sa se arate ca in orice triunghi \( ABC \) exista inegalitatile :

\( 1\odot\ \ \frac {1}{2R^2}\ \le\ \frac {1}{b^2+c^2}+\frac {1}{c^2+a^2}+\frac {1}{a^2+b^2} \) .

\( 2\odot\ \ \frac 12+\frac rR\ \le \frac {ab+bc+ca}{a^2+b^2+c^2}\ \le\ 1 \) .

Observatie. Puteti folosi prima inegalitate de aici.

1. Avem \( \sum\frac{1}{b^2+c^2}\ =\ \sum\frac{1}{4R^2(\sin^2 B+\sin^2 C)}\ =\ \sum\frac{1}{2R^2[2-(\cos 2B+cos 2C)]}\ \ge^{\mbox{C.B.S}}\ \frac{9}{2R^2(6-2\sum\cos 2A)}\ \ge^{(*)}\ \frac{9}{2R^2(6+3)}\ =\ \frac{1}{2R^2} \).

Am tinut cont de urmatoarea inegalitate: \( \cos 2A+\cos 2B+\cos 2C\ \ge\ -\frac{3}{2}\ (*) \) .

2.Vom merge prin echivalente
\( \odot\ \ \frac 12+\frac rR\ \le \frac {ab+bc+ca}{a^2+b^2+c^2} \)

\( \Leftrightarrow \) \( \frac{2(ab+bc+ca)-(a^2+b^2+c^2)}{a^2+b^2+c^2}\geq\frac{2r}{R} \)
\( \Leftrightarrow \) \( \frac{4r(4R+r)}{2p^2-8rR-2r^2}\geq\frac{2r}{R} \) \( \Leftrightarrow \) \( 4R^2+r^2+5Rr\geq p^2 \) (1)
Din Gerretsen si apoi Euler obtin succesiv \( p^2\leq4R^2+4Rr+3r^2\leq4R^2+4Rr+r^2+Rr=4R^2+r^2+5Rr \) adica inegalitatea (1)
P.S. Destul de "tare" inegalitatea de la 2.
Poate ne prezinta cineva o solutie mai eleganta:)

Se stie ca \( a^2+b^2+c^2\ \le\ 9R^2 \) . Asadar, \( \sum \frac {1}{b^2+c^2}\ \stackrel {C.B.S.}{\ge}\ \frac {9}{2(a^2+b^2+c^2)}\ \ge\ \frac {1}{2R^2} \) .

Incearcati sa aratati (dreapta !) ca \( \frac {1}{2R^2}\ \le\ \sum\frac {1}{b^2+c^2}\ \le\ \frac {1}{4Rr} \) .