`\qquad a;b;c>0=>a^3;b^3;c^3>0`
Áp dụng BĐT Cosi ta có:
`\qquad a+2b^3=a+b^3+b^3\ge 3`$\sqrt[3]{ab^6}=3\sqrt[3]{a}b^2$
`=>{2ab^3}/{a+2b^3}\le `$\dfrac {2\sqrt[3]{a}.b^2.b\sqrt[3]{a^2}}{3\sqrt[3]{a}b^2}=\dfrac{2}{3}b\sqrt[3]{a^2}$
Ta lại có:
$\quad a+a+1\ge 3\sqrt[3]{a^2}$
`=>`$\sqrt[3]{a^2}\le \dfrac{2a+1}{3}$
`=>2/3b`$\sqrt[3]{a^2}\le \dfrac{2(2a+1)b}{9}=\dfrac{4ab+2b}{9}$
`=>{2ab^3}/{a+2b^3}\le{4ab+2b}/9`
Tương tự chứng minh được:
`\qquad {2bc^3}/{b+2c^3}\le{4bc+2c}/9`
`\qquad {2ca^3}/{c+2a^3}\le {4ca+2a}/9`
Với mọi `a;b;c>0` ta có:
`\qquad (a-b)^2+(b-c)^2+(a-c)^2\ge 0`
`<=>a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\ge 0`
`<=>2(a^2+b^2+c^2)\ge 2(ab+bc+ac)`
`<=>a^2+b^2+c^2\ge ab+bc+ac`
`<=>a^2+b^2+c^2+2ab+2bc+2ac\ge 3(ab+bc+ac)`
`<=>(a+b+c)^2\ge 3(ab+bc+ac)`
`<=>ab+bc+ac\le {(a+b+c)^2}/3`
$\\$
`\qquad (2ab^3)/(a+2b^3)+(2bc^3)/(b+2c^3)+(2ca^3)/(c+2a^3)\le {4ab+2b}/9+{4bc+2c}/9+{4ca+2a}/9`
`\le 4/9 (ab+bc+ca)+2/9(a+b+c)`
`\le 4/9 {(a+b+c)^2}/3+2/9 (a+b+c)`
`\le 4/9 . {3^2}/3+2/9 . 3=2` (do `a+b+c=3)`
$\\$
Đẳng thức xảy ra khi: `a=b=c=1`
Vậy: `2\ge (2ab^3)/(a+2b^3)+(2bc^3)/(b+2c^3)+(2ca^3)/(c+2a^3)` với `a;b;c>0` thỏa `a+b+c=3`
