\(VT=\dfrac{a}{b\left(b^2+a\right)}-\dfrac{b}{c\left(c^2+b\right)}-\dfrac{c}{a\left(a^2+c\right)}\)
\(VT=\dfrac{a+b^2-b^2}{b\left(b^2+a\right)}-\dfrac{b+c^2-c^2}{c\left(c^2+b\right)}-\dfrac{c+a^2-a^2}{a\left(a^2+c\right)}\)
\(VT=\dfrac{1}{b}-\dfrac{b}{b^2+a}+\dfrac{1}{c}-\dfrac{c}{c^2+b}+\dfrac{1}{a}-\dfrac{a}{a^2+c}\)
\(VT=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\left(\dfrac{b}{b^2+a}+\dfrac{c}{c^2+b}+\dfrac{a}{a^2+c}\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\dfrac{b}{b^2+a}\ge\dfrac{b}{2b\sqrt{a}}=\dfrac{1}{2\sqrt{a}}\)
Thiết lập tương tự và thu lại tao có
\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\sqrt{\dfrac{1}{a}}+\sqrt{\dfrac{1}{b}}+\sqrt{\dfrac{1}{c}}\right)\)
Áp dụng bất đẳng thức Cauchy
\(\sqrt{\dfrac{1}{a}}+\sqrt{\dfrac{1}{b}}+\sqrt{\dfrac{1}{c}}\le\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3}{2}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\sqrt{\dfrac{1}{a}}+\sqrt{\dfrac{1}{b}}+\sqrt{\dfrac{1}{c}}\right)\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3\right)\)
\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3\right)\)
\(\Leftrightarrow VT\ge\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\)
Áp dụng bất đẳng thức Cauchy dạng phân thức
\(\Rightarrow\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\ge\dfrac{3}{4}.\dfrac{9}{a+b+c}+\dfrac{3}{4}=3\)
\(\Rightarrow VT\ge3\left(đpcm\right)\)
