Doua inegalitati in triunghi "made in Vietnam"

[mihai miculita]

\( \mbox{Sa se demonstreze ca intr-un triunghi oarecare ABC, au loc inegalitatile:} \)
a) \( \frac{a}{h_a+r_a}+\frac{b}{h_b+r_b}+\frac{c}{h_c+r_c}\ge\sqrt{3}; \)
b) \( \frac{a+b}{l_a+l_b}+\frac{a+c}{l_a+l_c}+\frac{b+c}{l_b+l_c}\ge2\sqrt{3}, \mbox{ unde }l_a, l_b, l_c \mbox{ sunt lungimile bisectoarelor.} \)

[Marius Mainea]

a)

b) \( LHS=\sum\frac{a+b}{\frac{2bc}{b+c}\sqrt{\frac{p(p-a)}{bc}}+\frac{2ca}{c+a}\sqrt{\frac{p(p-b)}{ca}}}\ge\sum\frac{a+b}{\sqrt{p(p-a)}+\sqrt{p(p-b)}}\ge \sum\frac{a+b}{\sqrt{p(p-a+p-b)\cdot 2}}\ge\frac{2\sqrt{ab}}{\sqrt{c(a+b+c)}}=\frac{2(ab+bc+ca)}{\sqrt{(a+b+c)abc}}\ge RHS \)

[mihai miculita]

a) \( \mbox{Facand substitutiile: } p-a=x, p-b=y, p-c=z, \mbox{ avem: } \)
\( a=y+z,\ S=\sqrt{xyz(x+y+z)},\ h_a=\frac{2S}{a}=\frac{2\sqrt{xyz(x+y+z)}}{y+z},\ r_a=\frac{S}{p-a}=\frac{\sqrt{xyz(x+y+z)}}{x}\Rightarrow \)
\( \Rightarrow \frac{a}{h_a+r_a}=\frac{y+z}{\(\frac{2}{y+z}+\frac{1}{x}\)\sqrt{xyz(x+y+z)}}=\frac{x(y+z)^2}{(2x+y+z)\sqrt{xyz.(x+y+z)}. \)
\( \mbox{Asa ca: }\sum{\frac{a}{h_a+r_a}\ge\sqrt{3}\Leftrightarrow \sum{\frac{x(y+z)^2}{2x+y+z}}}\ge\sqrt{3xyz(x+y+z)}\ (\forall)x,y,z>0. \)

voi reveni...