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Demonstrati ca in orice triunghi are loc inegalitatea
\( \frac{2(9R^2-p^2)}{9Rr}\ge\frac{\cos^2A}{\sin B \sin C}+\frac{\cos^2B}{\sin C \sin A}+\frac{\cos^2C}{\sin A \sin B}\ge 1 \).
I.V.Maftei si Dorel Baitan, Bucuresti

Andi Brojbeanu wrote:\( \triangle\ ABC\ \Longrightarrow\ \underline{\overline{\left\|\ 1\ \stackrel{(1)}{\le}\ \sum\ \frac{\cos^2 A}{\sin B\sin C}\ \stackrel{(2)}{\le}\ \frac{2(9R^2-p^2)}{9Rr}\ \right\|}} \) .

I.V.Maftei si Dorel Baitan, Bucuresti

\( \sum\ \frac{\cos^2 A}{\sin B\sin C}=\sum\ \frac{1-\sin^2 A}{\sin B\sin C}=\sum\ \frac{\sin A-\sin ^3 A}{\frac{rp}{2R^2}}=\frac{2R^2}{rp}\left(\frac pR-\frac {2p(p^2-6Rr-3r^2)}{8R^3}\right)=\frac{4R^2+6Rr+3r^2-p^2}{2Rr} \) .

Asadar, inegalitatea \( (1) \) este echivalenta cu inegalitatea lui Gerretsen i.e. \( \overline{\underline{\left\|\ p^2\ \le\ 4R^2+4Rr+3r^2\ \right\|}} \) .

Acum, inegalitatea \( (2) \) se reduce la : \( 9(4R^2+6Rr+3r^2-p^2)\ \le\ 4(9R^2-p^2)\ \Longleftrightarrow\ 54Rr+27r^2\ \le\ 5p^2 \) .

Comparand-o cu inegalitatea Gerretsen \( \underline{\overline{\left\|\ 16Rr-5r^2\ \le\ p^2\ \right\|}} \) ramane sa demonstram numai ca :

\( 54Rr+27r^2\ \le\ 5(16Rr-5r^2)\ \Longleftrightarrow\ 52r^2\ \le\ 26Rr\ \Longleftrightarrow\ 2r\ \le\ R\ \Longleftrightarrow\ \) inegalitatea lui Euler .

Observatie. Daca \( \triangle\ ABC \) este ascutitunghic avem inegalitatea : \( \underline{\overline{\left\|\ \sum\ \frac{\cos^2 A}{\sin B\sin C}\ \le\ \frac Rr-1\ \right\|}} \) .