Áp dụng bất đẳng thức Cauchy-Schwarz:
\(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=\dfrac{1}{a}+\dfrac{4}{2b}+\dfrac{9}{3c}\ge\dfrac{\left(1+2+3\right)^2}{a+2b+3c}=\dfrac{36}{a+2b+3c}\)
\(\dfrac{2}{a}+\dfrac{3}{b}+\dfrac{1}{c}=\dfrac{4}{2a}+\dfrac{9}{3b}+\dfrac{1}{c}\ge\dfrac{\left(2+3+1\right)^2}{2a+3b+c}=\dfrac{36}{2a+3b+c}\)
\(\dfrac{3}{a}+\dfrac{1}{b}+\dfrac{2}{c}=\dfrac{9}{3a}+\dfrac{1}{b}+\dfrac{4}{2c}\ge\dfrac{\left(3+1+2\right)^2}{3a+b+2c}=\dfrac{36}{3a+2b+c}\)
Cộng theo vế: \(6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge36F\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge6F\)
Mặt khác: \(ab+bc+ac=3abc\Leftrightarrow\dfrac{ab+bc+ac}{abc}=3\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
\(\Rightarrow18\ge36F\Leftrightarrow F\le\dfrac{1}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
