Sa se arate ca in orice triunghi \( ABC \) este adevarata inegalitatea
\( r\left(\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}\right)\leq R\left(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\right). \)
Inegalitatea 2, trigonometrica, cu sinA/2 si cos A/2
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Inegalitatea 2, trigonometrica, cu sinA/2 si cos A/2
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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maxim bogdan
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Se poate demonstra inegalitatea mai tare:
\( \frac{\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}}{\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}}<2\le \frac{R}{r} \)
Se considera un triunghi \( ABC \) si I centrul cercului inscris acestuia.
Vom folosi inegalitatea \( AI+BI>AB \)
In triunghiul ABI din teorema sinusului avem: \( \frac{BI}{\sin\frac{A}{2}}=\frac{AI}{\sin\frac{B}{2}}=\frac{AB}{\sin(\frac{\pi}{2}+\frac{C}{2})}=\frac{AB}{\cos\frac{C}{2}}=\frac{BI+AI}{\sin\frac{A}{2}+\sin\frac{B}{2}}>\frac{AB}{\sin\frac{A}{2}+\sin\frac{B}{2}} \)
Deci \( \sin\frac{A}{2}+\sin\frac{B}{2}>\cos\frac{C}{2} \).De aici e clar
\( \frac{\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}}{\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}}<2\le \frac{R}{r} \)
Se considera un triunghi \( ABC \) si I centrul cercului inscris acestuia.
Vom folosi inegalitatea \( AI+BI>AB \)
In triunghiul ABI din teorema sinusului avem: \( \frac{BI}{\sin\frac{A}{2}}=\frac{AI}{\sin\frac{B}{2}}=\frac{AB}{\sin(\frac{\pi}{2}+\frac{C}{2})}=\frac{AB}{\cos\frac{C}{2}}=\frac{BI+AI}{\sin\frac{A}{2}+\sin\frac{B}{2}}>\frac{AB}{\sin\frac{A}{2}+\sin\frac{B}{2}} \)
Deci \( \sin\frac{A}{2}+\sin\frac{B}{2}>\cos\frac{C}{2} \).De aici e clar
Feuerbach