\(\dfrac{a.h_a}{2}=S\Leftrightarrow a=\dfrac{2S}{h_a}\)
Tương tự:
\(b=\dfrac{2S}{h_b};c=\dfrac{2S}{h_c}\)
\(\dfrac{a+b+c}{4S}=\dfrac{\dfrac{2S}{h_a}+\dfrac{2S}{h_b}+\dfrac{2S}{h_c}}{4S}=\dfrac{2S\left(\dfrac{1}{h_a}+\dfrac{1}{h_b}+\dfrac{1}{h_c}\right)}{4S}=\dfrac{\dfrac{1}{h_a}+\dfrac{1}{h_b}+\dfrac{1}{h_c}}{2}\)
Tương đương:
\(\dfrac{1}{h_a+h_b}+\dfrac{1}{h_b+h_c}+\dfrac{1}{h_c+h_a}\le\dfrac{\dfrac{1}{h_a}+\dfrac{1}{h_b}+\dfrac{1}{h_c}}{2}\)
Cauchy-Schwarz:
\(\dfrac{1}{h_a+h_b}\le\dfrac{1}{4}\left(\dfrac{1}{h_a}+\dfrac{1}{h_b}\right)\)
\(\dfrac{1}{h_b+h_c}\le\dfrac{1}{4}\left(\dfrac{1}{h_b}+\dfrac{1}{h_c}\right)\)
\(\dfrac{1}{h_c+h_a}\le\dfrac{1}{4}\left(\dfrac{1}{h_c}+\dfrac{1}{h_a}\right)\)
Cộng theo vế suy ra đpcm