đặt A=[tex]\frac{1}{2+a^2b}+\frac{1}{2+b^2c}+\frac{1}{2+c^2a}[/tex]
2A=[tex]\frac{2}{2+a^2b}+\frac{2}{2+b^2c}+\frac{2}{2+c^2a}[/tex]
2A=[tex]\frac{2+a^2b-a^2b}{2+a^2b}+\frac{2+b^2c-b^2c}{2+b^2c}+\frac{2+c^2a-c^2a}{2+c^2a}[/tex]
2A=3 - [tex]\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}[/tex]
áp dụng bdt AM-GM
2+a^2b=1+1+a^2b[tex]\geq 3\sqrt[3]{a^2b}[/tex]
nên [tex]\frac{a^2b}{2+a^2b}\leq [tex]\frac{1}{3}[/tex] \sqrt[3]{a^4b^2}[/tex]
mặt khác [tex]\sqrt[3]{a^4b^2}=\sqrt[3]{ab.ab.a^2}\leq \frac{ab+ab+a^2}{3}[/tex]
[tex]\Leftrightarrow[/tex] [tex]\sqrt[3]{a^4b^2}\leq \frac{a^2+2ab}{3}[/tex]
Hay [tex]\frac{a^2b}{2+a^2b}\leq \frac{a^2+2ab}{9}[/tex]
tương tự [tex]\frac{b^2c}{2+b^2c}\leq \frac{b^2+2bc}{9}[/tex]
[tex]\frac{c^2a}{2+c^2a}\leq \frac{c^2+2ac}{9}[/tex]
do đó: 2A[tex]\geq 3- \frac{a^2+2ab+b^2+2bc+c^2+2ac}{9}[/tex]
2A[tex]\geq 3-\frac{(a+b+c)^2}{9}[/tex]
vì a+b+c=3 nên 2A[tex]\geq[/tex]2
A[tex]\geq[/tex]1
dấu bằng xảy ra :[tex]\Leftrightarrow[/tex]a=b=c=1