AD bđt AM-GM cho 3 số
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+C}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c}{a^3\left(b+c\right)}.\dfrac{\left(b+c\right)}{4bc}.\dfrac{1}{2b}}=\dfrac{3}{2a}\)
\(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}\ge\dfrac{3}{2a}-\dfrac{3}{4b}-\dfrac{1}{4c}\)
thiết lập bđt tương tự r cộng lại \(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\left(\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}\right)\left(a+b+c\right)=\dfrac{1}{2}\left(a+b+c\right)\)
bài này tui làm rùi :>
áp dụng BDT AM-GM cho 3 số thực dương
\(=>\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+c}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c.\left(b+c\right)}{a^3\left(b+c\right).2b}}=\dfrac{3}{2a}\)
\(=>\dfrac{b^2c}{a^3\left(b+c\right)}\ge\dfrac{3}{2a}-\dfrac{1}{4c}-\dfrac{1}{4b}-\dfrac{1}{2b}=\dfrac{3}{2a}-\dfrac{3}{4b}-\dfrac{1}{4c}\left(1\right)\)
tương tự \(=>\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{c+a}{4ac}+\dfrac{1}{2c}\ge\dfrac{3}{2b}\)
\(=>\dfrac{c^2a}{b^3\left(c+a\right)}\ge\dfrac{3}{2b}-\dfrac{1}{4a}-\dfrac{1}{4c}-\dfrac{1}{2c}=\dfrac{3}{2b}-\dfrac{3}{4c}-\dfrac{1}{4a}\)(2)
tương tự \(=>\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{a+b}{4ab}+\dfrac{1}{2a}\ge\dfrac{3}{2c}\)
\(=>\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{3}{2c}-\dfrac{1}{4b}-\dfrac{1}{4a}-\dfrac{1}{2a}=\dfrac{3}{2c}-\dfrac{3}{4a}-\dfrac{1}{4b}\left(3\right)\)
(1)(2)(3)
\(=>\)biểu thức đã cho đề bài \(\ge\)\(\dfrac{3}{2a}-\dfrac{3}{4b}-\dfrac{1}{4c}+\dfrac{3}{2b}-\dfrac{3}{4c}-\dfrac{1}{4a}+\dfrac{3}{2c}-\dfrac{3}{4a}-\dfrac{1}{4b}\)
\(=\left(\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)\(=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
:)) đến đây cứ sao sao ấy
