a) \(AH\perp BC\) \(\Rightarrow AH< AB;AH< AC\)
\(\Rightarrow2.AH< AB+AC\Leftrightarrow AH< \dfrac{AB+AC}{2}\)
b) Theo câu a ta có: \(AH< \dfrac{AB+AC}{2}\) \(\left(1\right)\)
Tương tự ta có: \(BK< \dfrac{AB+BC}{2}\) \(\left(2\right)\)
\(CI< \dfrac{CA+CB}{2}\) \(\left(3\right)\)
Từ \(\left(1\right)\),\(\left(2\right)\) và \(\left(3\right)\) \(\Rightarrow AH+BK+CI< AB+AC+BC\)