Inegalitatea 9, geometrica

[nica]

Sa se arate ca intr-un triunghi ascutitunghic are loc inegalitatea: \( \frac{1}{b^{2}+c^{2}-a^{2}}\ +\frac{1}{c^{2}+a^{2}-b^{2}}\ +\frac{1}{a^{2}+b^{2}-c^{2}}\geq\frac{1}{4r^{2}} \)

[Marian Dinca]

Solutie algebrica:
\( \mbox{Notand cu: }\\
b^2+c^2-a^2=x>0, \ c^2+a^2-b^2=y>0, \ a^2+b^2-c^2=z>0\Rightarrow a=\sqrt{\frac{y+z}{2}}, \ b=\sqrt{\frac{x+z}{2}}, \ c=\sqrt{\frac{x+y}{2}} \\
\mbox{avem }16S^2=\sum{a^2(b^2+c^2-a^2)}=\sum_{ciclic}{\frac{x(y+z)}{2}}=\sum_{ciclic}{xy} \mbox{ si } 4r^2=4.\frac{S^2}{r^2}={\frac{\sum{xy}}{(\sum\sqrt{\frac{x+y}{2}})^2}. \)
\(
\mbox{Asa ca inegalitatea de demonstrat este echivalenta cu: }\\
\sum{\frac{1}{x}}\ge \frac{1}{\frac{\sum{xy}}{(\sum{\sqrt{\frac{x+y}{2}}})^2}}=
\frac{(\sum{\sqrt{\frac{x+y}{2}}})^2}{\sum{xy}}\Leftrightarrow \sum{\frac{1}{x}}.\sum{xy}\ge(\sum{\sqrt{\frac{x+y}{2}}})^2\Leftrightarrow \sum{\frac{1}{x}}.\sum_{ciclic}{\frac{x(y+z)}{2}}\ge(\sum{\sqrt{\frac{y+z}{2}})^2
\)
\(
\mbox{si aplicam inegalitatea lui Cauchy-Buniakowski:}
\sum_{i=1}^3{a_i^2}.\sum_{i=1}^3{b_i^2}\ge(\sum_{i=1}^3{a_ib_i})^2.
\)

[nica]

Aceasta problema mai apare, sub o forma echivalenta, in revista Mathematical Reflections, nr. 4/2008 (problema O91), propusa de dl Profesor Mircea Becheanu. Ce sa zic...gand la gand cu bucurii...

[Virgil Nicula]

Sa se arate ca intr-un triunghi ascutitunghic are loc inegalitatea: \( \frac{1}{b^{2}+c^{2}-a^{2}}\ +\frac{1}{c^{2}+a^{2}-b^{2}}\ +\frac{1}{a^{2}+b^{2}-c^{2}}\geq\frac{1}{4r^{2}} \)

Metoda I. Reamintim relatia remarcabila intr-un triunghi \( ABC \) : \( \sum a\cdot\cos A=\frac {2pr}{R} \) .

Asadar, \( \sum\frac {1}{b^2+c^2-a^2}=\frac {1}{2abc}\cdot\sum \frac {a^2}{a\cdot\cos A}\ \stackrel{(C.B.S.)}{\ \ge\ }\ \frac {1}{2abc}\cdot\frac {\left(\sum a\right)^2}{\sum a\cos A}=\frac {1}{8Rpr}\cdot \frac {4p^2}{\frac {2pr}{R}}=\frac {1}{4r^2} \) .

Metoda II (similara !). Folosind identitatea \( \overline {\underline {\left|\ 4\cdot S=\left(b^2+c^2-a^2\right)\cdot\tan A\ \right|}} \) , inegalitatea este echivalenta cu

\( \overline {\underline {\left|\ \tan A+\tan B+\tan C\ge\frac pr\ \right|}} \) . Reamintim relatiile remarcabile intr-un triunghi \( ABC \) : \( \sum\sin A=\frac pR\ ; \)

\( \sum\sin 2A=4\prod\sin A=\frac {2pr}{R^2} \) . Asadar, \( \sum\tan A=2\cdot\sum \frac {\sin^2A}{\sin 2A}\ \stackrel{(C.B.S.)}{\ \ge\ }\ 2\cdot\frac {\left(\sum\sin A\right)^2}{\sum\sin 2A}=\frac {2\left(\frac pR\right)^2}{\frac {2pr}{R^2}}=\frac pr \) .

Observatie. Inegalitatea \( \sum \frac{1}{b^{2}+c^{2}-a^{2}}\ \stackrel{(C.B.S.)}{\ \ge\ }\ \frac {9}{a^2+b^2+c^2} \) este slaba deoarece \( \frac{1}{4r^{2}}\ge \frac {9}{a^2+b^2+c^2}\ (*) \) .

Intr-adevar, \( a^2+b^2+c^2=2\left(p^2-r^2-4Rr\right)\ge 2\left[\left(16Rr-5r^2\right)-r^2-4Rr\right]=12r(2R-r)\ge 36r^2 \) .

Problema propusa. Sa se arate ca intr-un triunghi ascutitunghic exista relatia

\( \overline{\underline{\left\|\ \frac {a}{b^2+c^2-a^2}+\frac {b}{c^2+a^2-c^2}+\frac {c}{a^2+b^2-c^2}\ \ge\ \frac {(a+b+c)^2}{3abc}\ \ge\ \frac {3(a+b+c)}{ab+bc+ca}\ \ge\ \frac {9}{a+b+c}\ \ge\ \frac {3(a+b+c)}{a^2+^2+c^2}\ \right\| \) .

[nica]

Avem: \( \sum\frac {a}{b^2+c^2-a^2}=\frac{1}{2abc}\sum\frac{a}{cosA}.a \) si aplicand Cebisev, obtinem:\( LHS\geq\frac{a+b+c}{3\cdot\2abc}\sum\frac{a}{cosA}=\frac{a+b+c}{3}\sum\frac{1}{2bc\cdot\ cosA}=\frac{a+b+c}{3}\sum\frac{1}{a^2+b^2-c^2}\geq RHS \). Am folosit inegalitatea :\( \sum\frac{1}{a^2+b^2-c^2}\geq\frac{9}{a^2+b^2+c^2} \)

[nica]

Interesant...Prima inegalitate se reduce la o cunoscuta inegalitate, si anume: \( \frac{3}{2}\geq \cos A+\cos B+\cos C \).

[nica]

Celelalte inegalitati se rezolva folosind inegalitatea dintre media aritmetica si cea geometrica ( inegalitatea 2) sau inegalitatea:\( \(a+b+c)^2\geq\3(ab+ac+bc) \) sau inca \( 3(a^2+b^2+c^2)\geq\(a+b+c)^2 \) (inegalitatea 3)