Toán 9 Bất đẳng thức

[tex]\frac{a^4b}{a^2+1}=\frac{a^2b(a^2+1)-a^2b}{a^2+1}=a^2b-\frac{a^2b}{a^2+1} \geq a^2b-\frac{a^2b}{2a}=a^2b-\frac{ab}{2}[/tex]
Lại có: [tex]a^2b+a^2b+c^2a\geq 3\sqrt[3]{a^5b^2c^2}=3a[/tex]
Tương tự [tex]b^2c+b^2c+a^2b\geq 3\sqrt[3]{b^5a^2c^2}=3b[/tex];[tex]c^2a+c^2a+b^2c\geq 3\sqrt[3]{a^2b^2c^5}=3c[/tex]
[tex]\Rightarrow a^2b+b^2c+c^2a\geq a+b+c[/tex]
Từ đó [tex]\frac{a^4b}{a^2+1}+\frac{b^4c}{b^2+1}+\frac{c^4a}{c^2+1}\geq a^2b+b^2c+c^2a-\frac{ab+bc+ca}{2}\geq \frac{1}{2}(a^2b+b^2c+c^2a+a+b+c-ab-bc-ca)\geq \frac{1}{2}(2ab+2bc+2ca-ab-bc-ca)=\frac{1}{2}(ab+bc+ca)\geq \frac{3}{2}\sqrt[3]{a^2b^2c^2}=\frac{3}{2}[/tex]