Doua perechi de inegalitati in triunghiuri particulare

[Mateescu Constantin]

Sa se arate ca in \( \triangle\ ABC \) avem : \( \ \ \ \begin{array}{cccc}\nearrow\ \left\|\ \begin{array}{cccc}
0\ <\ \phi\ <\ 90^{\circ} \\\\
A\ \ge\ B\ \ge\ \phi\ \ge\ C\end{array}\ \right\|\ \Longrightarrow\ \left\{\begin{array}{cccc}
a^2+b^2+c^2\ \ge\ 8R^2\cdot\sin^2\phi+4S\cdot\cot\phi \\\\
s\ \ge\ 2R\cdot\sin\phi+r\cdot\cot\frac{\phi}2\end{array} \\\\\\\\\\\\\\\\\\\\
\searrow\ \left\|\ \begin{array}{cccc}
0\ <\ \phi\ <\ 180^{\circ} \\\\
A\ \ge\ \phi\ \ge\ B\ \ge\ C\end{array}\ \right\|\ \Longrightarrow\ \left\{\begin{array}{cccc}
a^2+b^2+c^2\ \le\ 8R^2\cdot\sin^2\phi+4S\cdot\cot\phi \\\\
s\ \le\ 2R\cdot\sin\phi+r\cdot\cot\frac{\phi}2\end{array}\end{array} \)