Do \(ab+bc+ac=3abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
Áp dụng BĐT Cauchy cho 3 số \(\frac{1}{a};\frac{2}{b};\frac{3}{c}\) , ta có :
\(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=\frac{1}{a}+\frac{4}{2b}+\frac{9}{3c}\ge\frac{\left(1+2+3\right)^2}{a+2b+3c}=\frac{36}{a+2b+3c}\)
\(\Rightarrow\frac{1}{a+2b+3c}\le\frac{1}{36}\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right)\left(1\right)\)
CMTT , ta có : \(\frac{1}{2a+3b+c}\le\frac{1}{36}\left(\frac{2}{a}+\frac{3}{b}+\frac{1}{c}\right)\); \(\frac{1}{3a+b+2c}\le\frac{1}{36}\left(\frac{3}{a}+\frac{1}{b}+\frac{2}{c}\right)\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow F\le\frac{1}{36}\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}+\frac{2}{a}+\frac{3}{b}+\frac{1}{c}+\frac{3}{a}+\frac{1}{b}+\frac{2}{c}\right)\)
\(=\frac{1}{36}.6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{6}.3=\frac{1}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
Bạn Khôi Bùi làm đúng đó
vì sao có đc chổ (1) vậy?
Mình ko hiểu chổ đó.Bạn giải thích giùm mình đi.
\(\frac{1}{a+2b+c}\le\frac{1}{4}\left(\frac{1}{a+2b}+\frac{1}{3c}\right)\)\(\le\frac{1}{4}\left(\frac{1}{a+\frac{b}{2}}+\frac{1}{\frac{3b}{2}}\right)+\frac{1}{3c}\)\(\le\frac{1}{4}\left(\frac{1}{9a}+\frac{1}{9b}+\frac{1}{9c}\right)\)
Tương tự \(\hept{\begin{cases}\frac{1}{2a+3b+c}\le\frac{1}{4}\left(\frac{2}{9a}+\frac{1}{9b}+\frac{1}{9c}\right)\\\frac{1}{3a+b+2c}\le\frac{1}{4}\left(\frac{1}{3a}+\frac{1}{9b}+\frac{2}{9c}\right)\end{cases}}\)
=> \(F\le\frac{1}{4}\left(\frac{2}{3a}+\frac{2}{3b}+\frac{2}{3c}\right)=\frac{1}{6}\cdot\frac{ab+bc+ca}{abc}=\frac{1}{6}\cdot3=\frac{1}{2}\)
Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{2}\)
