Giải thích các bước giải:
Câu 3:
a.Ta có:
$S_{ABC}=\dfrac{1}{2}ah_a=\dfrac{1}{2}bh_b=\dfrac{1}{2}ch_c=\dfrac{3}{2}$
$\rightarrow ah_a=bh_b=ch_c=3$
$\rightarrow \begin{cases}\dfrac{1}{h_a}=\dfrac{a}{3}\\\dfrac{1}{h_b}=\dfrac{b}{3}\\\dfrac{1}{h_c}=\dfrac{c}{3}\end{cases}\rightarrow \dfrac{1}{h_a}+\dfrac{1}{h_b}+\dfrac{1}{h_c}=\dfrac{a+b+c}{3}$
b.Áp dụng bất đẳng thức bunhia cho 2 bộ số $(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}),(\dfrac{1}{h_a},\dfrac{1}{h_b},\dfrac{1}{h_c})$:
$(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})(\dfrac{1}{h_a}\dfrac{1}{h_b}\dfrac{1}{h_c})\ge (\sqrt[]{\dfrac{1}{a}.\dfrac{1}{h_a}}+\sqrt[]{\dfrac{1}{b}.\dfrac{1}{h_b}}+\sqrt[]{\dfrac{1}{c}.\dfrac{1}{h_c}})^2=3$
Dấu = xảy ra khi $a=b=c$
