Va ofer o inegalitate "mai tare" decat cunoscuta inegalitate \( \sum\cos A\le\frac 32 \) in orice \( ABC \) :
\( \underline {\overline {\left\|\ 12\cdot (\cos A + \cos B + \cos C)\le 15 + \cos (A - B) + \cos (B - C) + \cos (C - A)\ \right\|}}\ \le\ 18 \) .
Altfel spus, \( \underline {\overline {\left\|\ \frac {3(4r-R)}{R}\ \le\ \cos (A-B)+\cos (B-C)+\cos (C-A)\ \right\|}}\ \le\ 3 \) deoarece \( \sum\cos A=1+\frac rR \) .
O inegalitate trigonometrica "mai tare".
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Virgil Nicula
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Folosind formula lui Mollweide:
\( \frac{a+b}{c}=\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}} \)
avem:
\( RHS=15+\sum {(2\cos^2\frac{A-B}{2}-1)}\ge 12+6\sqrt[3]{\prod {\cos^2\frac{A-B}{2}}}\ge12+6\prod {\cos\frac{A-B}{2}}=12+6\prod {\frac{a+b}{c}\sin\frac{C}{2}}\ge12+6\prod {\frac{2\sqrt{ab}}{c}\sin\frac{C}{2}}=12+48\prod {\sin\frac{C}{2}}=12(1+4\prod {\sin\frac{C}{2}}) =LHS \)
\( \frac{a+b}{c}=\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}} \)
avem:
\( RHS=15+\sum {(2\cos^2\frac{A-B}{2}-1)}\ge 12+6\sqrt[3]{\prod {\cos^2\frac{A-B}{2}}}\ge12+6\prod {\cos\frac{A-B}{2}}=12+6\prod {\frac{a+b}{c}\sin\frac{C}{2}}\ge12+6\prod {\frac{2\sqrt{ab}}{c}\sin\frac{C}{2}}=12+48\prod {\sin\frac{C}{2}}=12(1+4\prod {\sin\frac{C}{2}}) =LHS \)