Aratati ca in triunghiul \( ABC \) au loc inegalitatile :
\( \bullet\ 2bc(1-\cos A) \ \le\ 2\left(1-\cos\frac{A}{3}\right)\left(b+2c\cos\frac{A}{3}\right)\left(c+2b\cos\frac{A}{3}\right)\ \le\ a^{2} \)
\( \bullet\ 2bc(1+\sin A) \ \le\ 2\left(1-\sin\frac{A}{3}\right)\left(b+2c\sin\frac{A}{3}\right)\left(c+2b\sin\frac{A}{3}\right) \ \le\ (b^{2}+c^{2})(1+\sin A)\ \)
Doua inegalitati in triunghi
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Mateescu Constantin
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Marius Mainea
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Notand \( x=\cos \frac{A}{2} \) si folosind relatia \( \cos A=4x^3-3x \) prima inegalitate este echivalenta cu
\( -4bcx^3+3xbc+bc\le -4bcx^3-2(b^2+c^2)x^2-xbc+4bcx^2+2(b^2+c^2)x+bc \)
sau
\( 0\le 2(b-c)^2x(1-x) \)
Pentru a doua folosind teorema cosinusului \( a^2=b^2+c^2-2b\cos A \) obtinem echivalent
\( 0\le (b-c)^2(2x-1)^2 \)
\( -4bcx^3+3xbc+bc\le -4bcx^3-2(b^2+c^2)x^2-xbc+4bcx^2+2(b^2+c^2)x+bc \)
sau
\( 0\le 2(b-c)^2x(1-x) \)
Pentru a doua folosind teorema cosinusului \( a^2=b^2+c^2-2b\cos A \) obtinem echivalent
\( 0\le (b-c)^2(2x-1)^2 \)