Bài làm :
\(VT=\frac{a}{b\left(b^2+a\right)}+\frac{b}{c\left(c^2+b\right)}+\frac{c}{a\left(a^2+c\right)}\)
\(=\frac{1}{b}\cdot\frac{a}{b^2+a}+\frac{1}{c}\cdot\frac{b}{c^2+b}+\frac{1}{a}\cdot\frac{c}{a^2+c}\)
\(=\frac{1}{b}\cdot\left(1-\frac{b^2}{b^2+a}\right)+\frac{1}{c}\cdot\left(1-\frac{c^2}{c^2+b}\right)+\frac{1}{a}\cdot\left(1-\frac{a^2}{a^2+c}\right)\)
Áp dụng BĐT Cô-si :
\(VT\ge\frac{1}{b}\cdot\left(1-\frac{b^2}{2b\sqrt{a}}\right)+\frac{1}{c}\cdot\left(1-\frac{c^2}{2c\sqrt{b}}\right)+\frac{1}{a}\cdot\left(1-\frac{a^2}{2a\sqrt{c}}\right)\)
\(=\frac{1}{b}\cdot\left(1-\frac{b}{2\sqrt{a}}\right)+\frac{1}{c}\cdot\left(1-\frac{c}{2\sqrt{b}}\right)+\frac{1}{a}\cdot\left(1-\frac{a}{2\sqrt{c}}\right)\)
\(=\frac{1}{b}-\frac{1}{2\sqrt{a}}+\frac{1}{c}-\frac{1}{2\sqrt{b}}+\frac{1}{a}-\frac{1}{2\sqrt{c}}\)
\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\cdot\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)
Lại áp dụng BĐT Cô-si :
\(\frac{1}{\sqrt{a}}\le\frac{\frac{1}{a}+1}{2};\frac{1}{\sqrt{b}}\le\frac{\frac{1}{b}+1}{2};\frac{1}{\sqrt{c}}\le\frac{\frac{1}{c}+1}{2}\)
Do đó :
\(VT\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\cdot\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}{2}\)
\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{4}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)
\(=\frac{3}{4}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\ge\frac{3}{4}\cdot\frac{9}{a+b+c}-\frac{3}{4}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
