Question

Sa se demonstreze ca in orice triunghi are loc relatia :
$l_{A} \leq \frac{b+c}{2}$

Answer 1

$\displaystyle In~orice~triunghi~este~cunoscuta~inegalitatea~dintre~lungimea \\ \\ inaltimii,~bisectoarei~si~a~medianei~dintr-un~varf: \\ \\ \boxed{h_a \le l_a \le m_a}~. \\ \\ Vom~demonstra~inegalitatea~m_a \le \frac{b+c}{2}~(chiar~daca~este~cunoscuta)$

$\displaystyle Consideram~triunghiul~ABC~si~punctul~M~-mijlocul~segmentului \\ \\ \left [BC \right].~Punctul~D~va~fi~simetricul~lui~A~fata~de~M. \\ \\ Patrulaterul~ACDB~este~paralelogram.~In~triunghiul~ACD~avem \\ \\ AD\ \textless \ AC+CD.~(inegalitatea~triunghiului) \\ \\ Dar~AD=2m_a~;~AC=b~;~CD=AB=c,~deci~2m_a\ \textless \ b+c. \\ \\ Deci~l_a \le m_a \ \textless \ \frac{b+c}{2}.$

$\displaystyle ------------ \\ \\ O ~alta~solutie:~l_a= \frac{2bc}{b+c} \cos \frac{A}{2}\ \textless \ \frac{b+c}{2} \cdot 1= \frac{b+c}{2}.$