Prove that
Median $AD>\frac{AB+AC-BC}{2}$
My attempt i tried using am gm inequality but cant figure out the way...
Any help will be appreciated..
Thank you
2026-05-15 21:27:02.1778880422
Prove that median $AD$ is greater than $\frac{AB+AC-BC}{2}$
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Applying triangle inequality
$AD+CD>AC$, $\triangle ACD$
$AD+BD>AB$, $\triangle ABD$
$2AD+CD+BD>AC+AB$
$2AD+BC>AC+AB$
$2AD>AC+AB-BC$