O inegalitate proprie intr-un triunghi ascutitunghic.

[Virgil Nicula]

Sa se arate ca daca triunghiul \( ABC \) este ascutitunghic atunci \( \overline{\underline{\left\|\sum\frac {a(a^2+2bc)}{b+c}\ \le\ \frac {4(R+r)^2(2R-r)}{R}\right\|}}\ \le\ 9R(2R-r). \)

[Mateescu Constantin]

Folosind teorema cosinusului partea stanga a inegalitatii este echivalenta cu :

\( \sum\ \frac{a(a^2+2bc)}{b+c}=\sum\ \frac{a(b^2+c^2-2bc\cos A+2bc)}{b+c}=\sum\ \frac{a[(b+c)^2-2bc\cos A]}{b+c}=\sum\ a(b+c)-2abc\sum\ \frac{\cos A}{b+c} \)

\( \Longleftrightarrow\ LHS=2(ab+bc+ca)-2abc\left\(\frac{\cos A}{b+c}+\frac{\cos B}{c+a}+\frac{\cos C}{c+a}\right\)\ (\ast) \)

In continuare voi demonstra inegalitatea (valabila in triunghi ascutitunghic !) : \( \overline{\underline{\left\|\ \sum\ \frac{\cos A}{b+c}\ \ge\ \frac{(R+r)^2}{2R^2p}\ \right\|}} \)

Intr-adevar , aplicand inegalitatea C.B.S. si folosind faptul ca \( \underline{\overline{\left\|\ \sum\ \cos A=1+\frac rR\ \right\|}} \) si \( \underline{\overline{\left\|\ a=b\cos C+c\cos B\ \right\|}} \) obtinem :

\( \sum\ \frac{\cos A}{b+c}=\sum\ \frac{\cos^2 A}{b\cos A+c\cos A}\ \ge^{\mbox{C.B.S.}}\ \frac{\left\(\sum\cos A\right\)^2}{\sum\ (b\cos A+c\cos A)}=\frac{\left\(1+\frac rR\right\)^2}{\sum\ (b\cos C+c\cos B)}=\frac{\frac{(R+r)^2}{R^2}}{\sum\ a}=\frac{(R+r)^2}{2R^2p} \)

Revenind in \( (\ast)\ \Longrightarrow\ LHS\le 2(ab+bc+ca)-8Rrp\ \cdot\ \frac{(R+r)^2}{2R^2p}=2(ab+bc+ca)-\frac{4r(R+r)^2}{R} \)

In continuare aplicam inegalitatea Gerretsen : \( \underline{\overline{\left\|\ ab+bc+ca\le 4(R+r)^2\ \right\|}}\ \Longrightarrow\ LHS\ \le\ 8(R+r)^2-\frac{4r(R+r)^2}{R} \)

\( \Longleftrightarrow\ \underline{\overline{\left\|\ \sum\ \frac{a(a^2+2bc)}{b+c}\ \le\ \frac{4(R+r)^2(2R-r)}R\ \right\|}}\ \ \mbox{\normal OK}\ \ \ \ \) Frumoasa inegalitate !

[Mateescu Constantin]

Va propun sa aratati urmatoarea inegalitate (partea stanga !) valabila in orice triunghi :

\( \underline{\overline{\left\|\ 2\[p^2-R(R-2r)\]\ \le\ \sum\ \frac{a(a^2+2bc)}{b+c}\ \right\|}} \) .