O inegalitate cu 1/cos(A/2) intr-un triunghi
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Mateescu Constantin
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O inegalitate cu 1/cos(A/2) intr-un triunghi
\( \fbox{\ \triangle\ ABC\ \ \Longrightarrow\ \ \frac 1{\cos\ \frac A2}\ +\ \frac 1{\cos\ \frac B2}\ +\ \frac 1{\cos\ \frac C2}\ \le\ \frac {\sqrt 3}2\ +\ \frac {3(4R+r)}{2s}\ } \)
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Cezar Lupu
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Acum vreo 2 ani am obtinut urmatoarea inegalitate (pe care inca am de gand s-o trimit la AMM),
\( \frac{4R+r}{s}+\sqrt{3}\leq\sum_{cyc}\frac{1}{\cos\frac{A}{2}}\leq\frac{2(4R+r)}{s}. \)
\( \frac{4R+r}{s}+\sqrt{3}\leq\sum_{cyc}\frac{1}{\cos\frac{A}{2}}\leq\frac{2(4R+r)}{s}. \)
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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Mateescu Constantin
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Cezar, partea dreapta la inegalitatea ta e mai slaba decat cea initiala, deoarece se reduce la \( s\sqrt 3\ \le\ 4R+r \) .
Pentru inegalitatea de inceput se aplica inegalitatea Popoviciu functiei \( f(x)=\tan\frac x2 \) , \( x\in (0,\pi) \) .
Totusi, partea stanga pare destul de tare ... Deocamdata am reusit sa o demonstrez in triunghiuri de tipul \( A\ge B\ge 60^{\circ}\ge C \) ...
Pentru inegalitatea de inceput se aplica inegalitatea Popoviciu functiei \( f(x)=\tan\frac x2 \) , \( x\in (0,\pi) \) .
Totusi, partea stanga pare destul de tare ... Deocamdata am reusit sa o demonstrez in triunghiuri de tipul \( A\ge B\ge 60^{\circ}\ge C \) ...
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Mateescu Constantin
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Hai ca-i usoara si partea stanga 
. Mai devreme m-am complicat eu inutil ...
Stim ca \( \fbox{\ \tan\frac x2=\frac {1-\cos x}{\sin x}\ } \) precum si \( \fbox{\ \sum\ \tan\ \frac A2=\frac {4R+r}s\ } \) .
Prin urmare avem : \( \sum\ \frac 1{\cos\ \frac A2}=\frac {4R+r}s\ +\ \sum\ \tan\ \frac {B+C}4 \) , deci ramane sa aratam ca :
\( \sum\ \tan\ \frac {\pi-A}4\ \ge\ sqrt 3 \) . Dar aceasta e tocmai inegalitatea Jensen
Asadar, am dovedit dubla inegalitate : \( \fbox{\ \frac {4R+r}s\ +\ \sqrt 3\ \le\ \sum\ \frac 1{\cos\ \frac A2}\ \le\ \frac {\sqrt 3}2\ +\ \frac {3(4R+r)}{2s}\ }\ \le\ \frac {2(4R+r)}s \) .
. Mai devreme m-am complicat eu inutil ...Stim ca \( \fbox{\ \tan\frac x2=\frac {1-\cos x}{\sin x}\ } \) precum si \( \fbox{\ \sum\ \tan\ \frac A2=\frac {4R+r}s\ } \) .
Prin urmare avem : \( \sum\ \frac 1{\cos\ \frac A2}=\frac {4R+r}s\ +\ \sum\ \tan\ \frac {B+C}4 \) , deci ramane sa aratam ca :
\( \sum\ \tan\ \frac {\pi-A}4\ \ge\ sqrt 3 \) . Dar aceasta e tocmai inegalitatea Jensen
Asadar, am dovedit dubla inegalitate : \( \fbox{\ \frac {4R+r}s\ +\ \sqrt 3\ \le\ \sum\ \frac 1{\cos\ \frac A2}\ \le\ \frac {\sqrt 3}2\ +\ \frac {3(4R+r)}{2s}\ }\ \le\ \frac {2(4R+r)}s \) .