Sa se arate ca in \( \triangle\ ABC \) avem \( \begin{array}{cccc}
\nearrow\ 1.\ & A=30^{\circ} & \Longleftrightarrow & b^2+c^2=a^2+4S\sqrt 3.\\\\\\\\
\searrow\ 2.\ & A\in\left(\ 0\ ,\ \frac {\pi}{6}\ \right] & \Longrightarrow & b^2+c^2\ \ge\ a^2+4S\sqrt 3.\end{array} \)
O echivalenta si o implicatie intr-un triunghi ABC.
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Virgil Nicula
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1) \( b^2+c^2=a^2+4S\sqrt 3 \) \( \Longleftrightarrow \) \( 2bc\cos A=4\frac{bc\sin A}{2}\sqrt{3} \) \( \Longleftrightarrow \) \( \tan A=\frac{1}{\sqrt{3}} \) \( \Longleftrightarrow \) \( A=30^{\circ}. \)
2) \( A\in\left(\ 0\ ,\ \frac {\pi}{6}\ \right] \) \( \Longrightarrow \) \( \tan A\le\frac{1}{\sqrt{3}} \) \( \Longrightarrow \) \( b^2+c^2\ \ge\ a^2+4S\sqrt 3. \)
2) \( A\in\left(\ 0\ ,\ \frac {\pi}{6}\ \right] \) \( \Longrightarrow \) \( \tan A\le\frac{1}{\sqrt{3}} \) \( \Longrightarrow \) \( b^2+c^2\ \ge\ a^2+4S\sqrt 3. \)