Inegalitate trigonometrica

[Claudiu Mindrila]

Daca \( A,\ B,\ C \) sunt masurile unghiurilor unui triunghi, atunci \( \cos\frac{A-B}{2}+\cos\frac{B-C}{2}+\cos\frac{C-A}{2}\ge\sin\frac{3A}{2}+\sin\frac{3B}{2}+\sin\frac{3C}{2} \).

[Theodor Munteanu]

\( \sin\frac{3A}{2}+\sin\frac{3B}{2}+\sin\frac{3C}{2}=\cos(\frac{\pi}{2}-\frac{3A}{2})+\cos(\frac{\pi}{2}-\frac{3B}{2}-\frac{3C}{2})=\cos(\frac{B+C-2A}{2})+\cos(\frac{A+C-2B}{2})+\cos(\frac{A+B-2C}{2})=\cos(\frac{B-A+C-A}{2})+\cos(\frac{A-B+C-B}{2})+\cos(\frac{A-C+B-C}{2}) \)
\( \cos(\frac{B-A+C-A}{2}\le\frac{cos(B-A)+cos(C-A)}{2} \) si analoagele si adunandu-le obtinem inegalitatea de demonstrat \( \cos\frac{A-B}{2}+\cos\frac{B-C}{2}+\cos\frac{C-A}{2}\ge cos(A-B)+cos(B-C)+cos(C-A) \rightarrow \cos(A-B)-\cos(\frac{A-B}{2})+\cos(B-C)-\cos(\frac{B-C}{2})+\cos(C-A)-\cos(\frac{C-A}{2})\le 0 \), ceea ce e evident deoarece functia cos e descrescatoare.

[Mateescu Constantin]

\( \cos\frac {A-B}{2}=\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{A}{2}\sin\frac{B}{2}=\sqrt{\frac{p^{2}(p-a)(p-b)}{abc^{2}}}+\sqrt{\frac{(p-a)(p-b)(p-c)^{2}}{abc^{2}}}. \)

\( \sum\cos\frac{A-B}{2}=\sum\frac{a+b}{c}\sqrt{\frac{(p-a)(p-b)}{ab}}=\sum\frac{a+b}{c}\cdot\sin\frac{C}{2}, \) deci ramane de aratat ca:

\( \sum\frac{a+b}{c}\cdot\sin\frac{C}{2}\geq3\sum\sin\frac{A}{2}-4\sum\sin^{3}\frac{A}{2} \)

\( \sum\left\(\frac{a+b-3c}{c}+\frac{4(p-a)(p-b)}{ab}\right\)\cdot\sin\frac{C}{2}\geq0 \)

\( \sum\frac{(b-c)(a-c)(a+b+c)}{abc}\cdot\sin\frac{C}{2}\geq0 \)

\( \Leftrightarrow \sum(c-b)(c-a)\cdot\sin\frac{C}{2}\geq0. \)

Presupunem, fara a restrange generalitatea ca \( a\geq b\geq c. \)

Atunci \( (c-b)(c-a)\cdot\sin\frac{C}{2}\geq0 \) si ne mai ramane de demonstrat ca:

\( (b-a)(c-a)\cdot\sin\frac{A}{2}+(a-b)(c-b)\cdot\sin\frac{B}{2}\geq0 \)

\( \Leftrightarrow (a-b)\left\[(a-c)\cdot\sin\frac{A}{2}-(b-c)\cdot\sin\frac{B}{2}\right\]\geq0 \) evident, deoarece \( A\geq B \), si deci \( \sin\frac{A}{2}\geq \sin\frac{B}{2}\ si\ a-c\geq b-c. \)