Có: `8a^3+1=(2a+1)(4a^2-2a+1)<=((4a^2+2)/2)^2=(2a^2+1)^2`
`=> \sqrt{8a^3+1}<=2a^2+1`
CMTT:....
`=> \sum a^2/(\sqrt((8a^3+1)(8b^3+1)))>=\sum a^2/((2a^2+1)(2b^2+1))`
Cần cm: `\sum a^2/((2a^2+1)(2b^2+1))>=1/3`
`<=>3sum ((a^2)(2c^2+1))>= (2a^2+1)(2b^2+1)(2c^2+1)`
`<=>6\sum a^2c^2+3\sum a^2>= 8a^2b^2c^2+4\suma^2c^2+2\sum a^2+1`
`<=>2\sum a^2c^2+\sum a^2>=9`
Mà theo cauchy: `\sum a^2c^2>=3` $\sqrt[3]{a^4b^4c^4}=3$
`\sum a^2>=3` $\sqrt[3]{a^2b^2c^2}=3$
`=>2\sum a^2c^2+\sum a^2>=9` (đpcm)
Dấu `=` xảy ra khi : `a=b=c=1`