Prove that $AH<\frac{AB+AC}{2}$ in a triangle

Question

Prove that $AH<\frac{AB+AC}{2}$ in $\triangle ABC$. AH is Height of triangle from A to BC.

I thought it can be done by squaring the sides but I didn't get to any good result.

Thanks and sorry for my English.

Answer 1

$AH<AB$ and $AH<AC$. Then $$2AH<AB+AC$$ Then $$AH<\frac{AB+AC}2$$