Question Number 199481 by tri26112004 last updated on 04/Nov/23
$${Give}\:\bigtriangleup{ABC}\:{is}\:{acute}\:{triangle}. \\ $$$${M}\:\:{is}\:{a}\:{midpoint}\:{of}\:{BC} \\ $$$${Prove}\:{that}\:{AB}+{AC}>\mathrm{2}{AM} \\ $$
Answered by mr W last updated on 04/Nov/23
Commented by mr W last updated on 04/Nov/23
$${AB}+{BA}'>{AA}' \\ $$$$\Rightarrow{AB}+{AC}>\mathrm{2}{AM} \\ $$
Commented by tri26112004 last updated on 04/Nov/23
$${you}\:{have}\:{other}\:{solution}¿ \\ $$
Commented by mr W last updated on 04/Nov/23
$${this}\:{is}\:{the}\:{simplest}\:{solution}. \\ $$
Commented by mr W last updated on 04/Nov/23
$${an}\:{other}\:{solution}: \\ $$$${BC}+{AC}>{AB}\:\Rightarrow{BC}>{AB}−{AC} \\ $$$$\mathrm{2}{AM}=\sqrt{\mathrm{2}{AB}^{\mathrm{2}} +\mathrm{2}{AC}^{\mathrm{2}} −{BC}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:<\sqrt{\mathrm{2}{AB}^{\mathrm{2}} +\mathrm{2}{AC}^{\mathrm{2}} −\left({AB}−{AC}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\left({AB}+{AC}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={AB}+{AC} \\ $$
Commented by tri26112004 last updated on 05/Nov/23
$${Good}\:{solution} \\ $$