Inegalitate clasica (de manual ... vechi !).

[Virgil Nicula]

Sa se arate ca in orice triunghi \( ABC \) exista relatia \( \overline {\underline {\left\|\ \sqrt {\frac {2r}{R}}\ \le\ \cos\frac {B-C}{2}\ \right\|}}\ \ (*) \)

si se atinge egalitatea daca si numai daca \( b+c=2a \).

[Marius Mainea]

Folosim relatia \( r=4\sin \frac{A}{2}\sin \frac{B}{2}\sin\frac{C}{2} \) si inegalitatea este echivalenta cu

\( 8\sin \frac{A}{2}\sin \frac{B}{2}\sin\frac{C}{2}\le \cos^2\frac{B-C}{2} \) sau

\( 4\sin\frac{A}{2}(\cos\frac{B-C}{2}-\sin\frac{A}{2})\le \cos^2\frac{B-C}{2} \) sau

\( (\cos\frac{B-C}{2}-2\sin\frac{A}{2})^2\ge 0 \)

[red_dog]

La fel, folosind identitatea \( r=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{B}{2} \) inegalitatea devine

\( \sqrt{8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}\leq \cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{B}{2}\sin\frac{C}{2} \)

Apoi, folosind formulele
\( \sin\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{bc}} \) si celelalte obtinute prin permutari circulare si \( \cos\frac{B}{2}=\sqrt{\frac{p(p-b)}{ac}} \) si cea analoaga pentru ubghiul B si facand calculele, inegalitatea este echivalenta cu

\( \frac{b+c}{a}\geq 2\sqrt{\frac{b+c}{a}-1}\Leftrightarrow\left(\frac{b+c}{a}-2\right)^2\geq 0 \)