Ta có : \(\Sigma\dfrac{ab}{a^2+b^2}=3-\Sigma\dfrac{a^2+b^2-ab}{a^2+b^2}\)
Thấy : \(0< ab\left(a^2+b^2-ab\right)\le\dfrac{\left(a^2+b^2\right)^2}{4}\)
\(\Rightarrow\dfrac{a^2+b^2-ab}{a^2+b^2}\le\dfrac{1}{4}\left(\dfrac{a^2+b^2}{ab}\right)=\dfrac{1}{4}\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)
CMTT ; ta có : \(\dfrac{b^2+c^2-bc}{b^2+c^2}\le\dfrac{1}{4}\left(\dfrac{b}{c}+\dfrac{c}{b}\right);\dfrac{c^2+a^2-ac}{a^2+c^2}\le\dfrac{1}{4}\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)
Suy ra : \(\Sigma\dfrac{ab}{a^2+b^2}\ge3-\dfrac{1}{4}\left(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}\right)=\dfrac{1}{4}\left(\dfrac{a+c}{b}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\right)\)
Thấy : \(\dfrac{a+c}{b}+\dfrac{b+c}{a}+\dfrac{a+b}{c}=\dfrac{\left(a+c\right)ac+\left(b+c\right)bc+ab\left(a+b\right)}{abc}=ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\)( do abc = 1 )
Áp dụng BĐT Schur ta được : \(ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\le a^3+b^3+c^3+3abc=\Sigma a^3+3\)
Suy ra : \(\Sigma\dfrac{ab}{a^2+b^2}\ge3-\dfrac{1}{4}\left(\Sigma a^3+3\right)=\dfrac{9}{4}-\dfrac{1}{4}\Sigma a^3\cdot\)
Khi đó : \(\Sigma a^3+\Sigma\dfrac{ab}{a^2+b^2}\ge\dfrac{3}{4}\Sigma a^3+\dfrac{9}{4}\ge\dfrac{3}{4}.3+\dfrac{9}{4}=\dfrac{9}{2}\)
" = " <=> a = b = c = 1
Vậy ...
Khuya rồi còn đăng à bạn ?
