Sa se arate ca intr-un triunghi \( ABC \) exista echivalenta
\( \underline{\overline{\left\|\ \cos\ (B-C)\ \le\ \frac {2bc}{b^2+c^2} \ \Longleftrightarrow\ A\ \le\ 90^{\circ}\ \ \vee\ \ B=C\ \Longleftrightarrow\ \frac rR\ \le\ \frac {a(b+c-a)}{b^2+c^2}\ \Longleftrightarrow\ \frac {2a(p-b)(p-c)}{p\left(b^2+c^2\right)}\ \le\ \frac rR\ \right\|}} \) .
Observatie. Se arata usor ca in orice triunghi \( \frac {a(b+c-a)}{b^2+c^2}\ \le\ \frac 12 \) .
Inegalitate intr-un triunghi.
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Virgil Nicula
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Inegalitate intr-un triunghi.
Last edited by Virgil Nicula on Fri Aug 07, 2009 7:52 pm, edited 1 time in total.
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Marius Mainea
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Folosim \( \cos B=\frac{a^2+c^2-a^2}{2ac} \) \( \sin B=\frac{2S}{ac} \)
\( \odot\ \cos\ (B-C)\ \le\ \frac {2bc}{b^2+c^2}\Longleftrightarrow
\frac{a^2+c^2-a^2}{2ac}\cdot\frac{a^2+b^2-c^2}{2ab}+\frac{4S^2}{a^2bc}\le\
frac{2bc}{b^2+c^2}\Longleftrightarrow a^2(b^2-c^2)\le (b^2+c^2)(b^2-c^2)\Longleftrightarrow\angle{A}\le 90^{\circ}\vee\ \ B=C\ \)
\( \odot\frac rR\ \le\ \frac {a(b+c-a)}{b^2+c^2}\Longleftrightarrow \frac{4R\sin \frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{R}\le \frac{2R2\sin\frac{A}{2}\cos\frac{A}{2}\cdot 2R\cdot 4\cos\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{4R^2(\sin^2 B+\sin^2 C)}\Longleftrightarrow \sin^2 B+\sin^2 C\le 2\cos^2\frac{A}{2}\Longleftrightarrow\cos A\cos (B-C)\le \cos A\Longleftrightarrow A\ \le\ 90^{\circ}\ \ \vee\\ B=C\ \)
\( \odot\ \cos\ (B-C)\ \le\ \frac {2bc}{b^2+c^2}\Longleftrightarrow
\frac{a^2+c^2-a^2}{2ac}\cdot\frac{a^2+b^2-c^2}{2ab}+\frac{4S^2}{a^2bc}\le\
frac{2bc}{b^2+c^2}\Longleftrightarrow a^2(b^2-c^2)\le (b^2+c^2)(b^2-c^2)\Longleftrightarrow\angle{A}\le 90^{\circ}\vee\ \ B=C\ \)
\( \odot\frac rR\ \le\ \frac {a(b+c-a)}{b^2+c^2}\Longleftrightarrow \frac{4R\sin \frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{R}\le \frac{2R2\sin\frac{A}{2}\cos\frac{A}{2}\cdot 2R\cdot 4\cos\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{4R^2(\sin^2 B+\sin^2 C)}\Longleftrightarrow \sin^2 B+\sin^2 C\le 2\cos^2\frac{A}{2}\Longleftrightarrow\cos A\cos (B-C)\le \cos A\Longleftrightarrow A\ \le\ 90^{\circ}\ \ \vee\\ B=C\ \)
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Virgil Nicula
- Euler
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\( \underline{\overline{\left\|\ \cos (B-C)\le \frac {2bc}{b^2+c^2}\ \right\|}} \) . Adunam stanga/dreapta \( 1 \) si folosim relatia \( 1+\cos x=2\cos^2\frac x2 \) .
\( \Longleftrightarrow\ \left\|\ 2\cos^2\frac {B-C}{2}\le \frac {(b+c)^2}{b^2+c^2}\ \right\| \) . Inmultim cu \( 2\sin ^2\frac {B+C}{2}=2\cos^2\frac A2>0 \) .
Folosim relatiile \( 2\sin\frac {B+C}{2}\cos\frac {B-C}{2}=\sin B+\sin C \) si \( \cos^2\frac A2=\frac {p(p-a)}{bc} \) .
\( \Longleftrightarrow\ \left\|\ (\sin B+\sin C)^2\le \frac {(b+c)^2}{b^2+c^2}\cdot\frac {2p(p-a)}{bc}\ \right\| \) . Insa \( \sin B=\frac {b}{2R} \) , \( \sin C=\frac {c}{2R} \) .
\( \Longleftrightarrow\ \left\|\ bc\left(b^2+c^2\right)\le 4R^2\cdot 2p(p-a)\ \right\| \) . Inmultim cu \( r \) si folosim relatia \( 4Rpr=abc \) .
\( \Longleftrightarrow\ \left\|\ rbc\left(b^2+c^2\right)\le R\cdot abc\cdot 2p(p-a)\ \right\| \) . Simplificam prin \( bc>0 \) si stim ca \( 2(p-a)=b+c-a \) .
\( \Longleftrightarrow\ \underline{\overline{\left\|\ \frac rR\le\frac {a(b+c-a)}{b^2+c^2}\ \right\|}} \) \( \Longleftrightarrow\ \frac {4(p-a)(p-b)(p-c)}{abc}\le \frac {2a(p-a)}{b^2+c^2} \)
\( \Longleftrightarrow\ 2\left(b^2+c^2\right)(p-b)(p-c)\le a^2bc \) \( \Longleftrightarrow\ \left(b^2+c^2\right)\left[a^2-(b-c)^2\right]\le 2a^2bc \)
\( \Longleftrightarrow\ a^2\left[\left(b^2+c^2\right)-2bc\right]\le\left(b^2+c^2\right)(b-c)^2 \) \( \Longleftrightarrow\ a^2(b-c)^2\le \left(b^2+c^2\right)(b-c)^2 \)
\( \Longleftrightarrow\ \underline{\overline{\left\|\ B=C\ \ \vee\ \ A\ \le\ 90^{\circ}\ \right\|}} \) deoarece \( a^2\le b^2+c^2\ \Longleftrightarrow\ A\ \le\ 90^{\circ} \) .
\( \Longleftrightarrow\ \left\|\ 2\cos^2\frac {B-C}{2}\le \frac {(b+c)^2}{b^2+c^2}\ \right\| \) . Inmultim cu \( 2\sin ^2\frac {B+C}{2}=2\cos^2\frac A2>0 \) .
Folosim relatiile \( 2\sin\frac {B+C}{2}\cos\frac {B-C}{2}=\sin B+\sin C \) si \( \cos^2\frac A2=\frac {p(p-a)}{bc} \) .
\( \Longleftrightarrow\ \left\|\ (\sin B+\sin C)^2\le \frac {(b+c)^2}{b^2+c^2}\cdot\frac {2p(p-a)}{bc}\ \right\| \) . Insa \( \sin B=\frac {b}{2R} \) , \( \sin C=\frac {c}{2R} \) .
\( \Longleftrightarrow\ \left\|\ bc\left(b^2+c^2\right)\le 4R^2\cdot 2p(p-a)\ \right\| \) . Inmultim cu \( r \) si folosim relatia \( 4Rpr=abc \) .
\( \Longleftrightarrow\ \left\|\ rbc\left(b^2+c^2\right)\le R\cdot abc\cdot 2p(p-a)\ \right\| \) . Simplificam prin \( bc>0 \) si stim ca \( 2(p-a)=b+c-a \) .
\( \Longleftrightarrow\ \underline{\overline{\left\|\ \frac rR\le\frac {a(b+c-a)}{b^2+c^2}\ \right\|}} \) \( \Longleftrightarrow\ \frac {4(p-a)(p-b)(p-c)}{abc}\le \frac {2a(p-a)}{b^2+c^2} \)
\( \Longleftrightarrow\ 2\left(b^2+c^2\right)(p-b)(p-c)\le a^2bc \) \( \Longleftrightarrow\ \left(b^2+c^2\right)\left[a^2-(b-c)^2\right]\le 2a^2bc \)
\( \Longleftrightarrow\ a^2\left[\left(b^2+c^2\right)-2bc\right]\le\left(b^2+c^2\right)(b-c)^2 \) \( \Longleftrightarrow\ a^2(b-c)^2\le \left(b^2+c^2\right)(b-c)^2 \)
\( \Longleftrightarrow\ \underline{\overline{\left\|\ B=C\ \ \vee\ \ A\ \le\ 90^{\circ}\ \right\|}} \) deoarece \( a^2\le b^2+c^2\ \Longleftrightarrow\ A\ \le\ 90^{\circ} \) .