Biến đổi :
\(VT=\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ca}=\frac{a}{b\left(a+b^2\right)}+\frac{b}{c\left(b+c^2\right)}+\frac{c}{a\left(c+a^2\right)}\)
\(=\frac{1}{b}\cdot\frac{a}{a+b^2}+\frac{1}{c}\cdot\frac{b}{b+c^2}+\frac{1}{a}\cdot\frac{1}{c+a^2}\)
\(=\frac{1}{b}\cdot\left(1-\frac{b^2}{a+b^2}\right)+\frac{1}{c}\cdot\left(1-\frac{c^2}{b+c^2}\right)+\frac{1}{a}\cdot\left(1-\frac{a^2}{c+a^2}\right)\)
Áp dụng BĐT Cô-si :
\(VT\ge\frac{1}{b}\cdot\left(1-\frac{b^2}{2b\sqrt{a}}\right)+\frac{1}{c}\cdot\left(1-\frac{c^2}{2c\sqrt{b}}\right)+\frac{1}{a}\cdot\left(1-\frac{a^2}{2a\sqrt{c}}\right)\)
\(=\frac{1}{b}-\frac{1}{2\sqrt{a}}+\frac{1}{c}-\frac{1}{2\sqrt{b}}+\frac{1}{a}-\frac{1}{2\sqrt{c}}\)
\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\cdot\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)
Áp dụng BĐT quen thuộc : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) và BĐT Cô-si ta có:
\(VT\ge\frac{9}{a+b+c}-\frac{1}{2}\cdot\left(\frac{\frac{1}{a}+1}{2}+\frac{\frac{1}{b}+1}{2}+\frac{\frac{1}{c}+1}{2}\right)\)
\(=\frac{9}{3}-\frac{1}{2}\cdot\left(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}{2}\right)\ge3-\frac{1}{2}\cdot\left(\frac{\frac{9}{a+b+c}+3}{2}\right)\)
\(=3-\frac{1}{2}\cdot\left(\frac{\frac{9}{3}+3}{2}\right)=\frac{3}{2}\)
Ta có đpcm.
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
