mateforum.ro

\mbox{Notam cu: } x=S_{PBC}, y=S_{PCA}, z=S_{PAB}.\\ \mbox{Cu aceste notatii au loc relatiile: } ad_a=2x, \ bd_b=2y, \ cd_c=2z \mbox{ si } ah_a+bh_b+ch_c=2(x+y+z). \\ \mbox{Asa ca inegalitatea de demonstrat este echivalenta in mod succesiv cu: }\\ \sum{d_ah_a^2\ge \left(\sum{d_a}\right)^3\Leftright...

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{...