Cho a,b,c>0 và a+b+c=3CMR
\(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ac}\ge\frac{3}{2}\)
Áp dụng BĐT AM-GM ta có:
\(VT=\dfrac{1}{a}-\dfrac{a}{c+a^2}+\dfrac{1}{b}-\dfrac{b}{a+b^2}+\dfrac{1}{c}-\dfrac{c}{b+c^2}\)
\(=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\left(\dfrac{a}{c+a^2}+\dfrac{b}{a+b^2}+\dfrac{c}{b+c^2}\right)\)
\(\ge\dfrac{9}{a+b+c}-\left(\dfrac{a}{2a\sqrt{c}}+\dfrac{b}{2b\sqrt{a}}+\dfrac{c}{2c\sqrt{b}}\right)\)
\(\ge3-\left(\dfrac{1}{2\sqrt{c}}+\dfrac{1}{2\sqrt{a}}+\dfrac{1}{2\sqrt{b}}\right)\)\(=3-\left(\dfrac{2\sqrt{a}}{4a}+\dfrac{2\sqrt{b}}{4b}+\dfrac{2\sqrt{c}}{4c}\right)\)
\(\ge3-\left(\dfrac{a+1}{4a}+\dfrac{b+1}{4b}+\dfrac{c+1}{4c}\right)\)
\(=3-\left(\dfrac{3}{4}+\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}\right)\ge3-\left(\dfrac{3}{4}+\dfrac{9}{4\left(a+b+c\right)}\right)=\dfrac{3}{2}\)
Khi \(a=b=c=1\)
