ab+bc+ca=1
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Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
ab+bc+ca=1
\( a,\ b,\ c\ >\ 0,\ ab+bc+ca=1\ \Longrightarrow\ \sum\frac{1-a^2}{1+a^2}\ \le\ \frac 32 \).
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Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Solutia 1)
\( LHS=\sum \frac{1+a^2-2a^2}{1+a^2}=3-2\sum\frac{a^2}{1+a^2}\le3-2\frac{(a+b+c)^2}{3+a^2+b^2+c^2}\le 3-\frac{3}{2}=RHS \)
Solutia 2) Notam \( a=\tan \frac{A}{2},b=\tan\frac{B}{2},c=\tan\frac{C}{2} \) si inegalitatea devine
\( \sum\cos A\le\frac{3}{2} \) care este adevarata deoarece cos e concava pe \( [0,\pi] \)
\( LHS=\sum \frac{1+a^2-2a^2}{1+a^2}=3-2\sum\frac{a^2}{1+a^2}\le3-2\frac{(a+b+c)^2}{3+a^2+b^2+c^2}\le 3-\frac{3}{2}=RHS \)
Solutia 2) Notam \( a=\tan \frac{A}{2},b=\tan\frac{B}{2},c=\tan\frac{C}{2} \) si inegalitatea devine
\( \sum\cos A\le\frac{3}{2} \) care este adevarata deoarece cos e concava pe \( [0,\pi] \)