`6.`

Ta có:

`(a-1)^2>=0`

`a^2-2a+1>=0`

`4a^2+4a+4>=3a^2+6a+3`

`4(a^2+a+1)>=3(a+1)^2`

`((a+1)^2)/(a^2+a+1)<=4/3`

`(a+1)/(sqrt(a^2+a+1))<=(2sqrt(3))/3`

`(b+1)/(sqrt(b^2+b+1))<=(2sqrt(3))/3`

`(c+1)/(sqrt(c^2+c+1))<=(2sqrt(3))/3`

`=> P<=(2sqrt(3))/3+(2sqrt(3))/3+(2sqrt(3))/3=2sqrt(3)`

Dấu "=" xảy ra khi `a=b=c=1`

Vậy GTLN của `P` là `2sqrt(3)`

`7.`

Ta có:

`P=(a^4)/(asqrt(b^2+1))+(b^4)/(bsqrt(c^2+1))+(c^4)/(csqrt(a^2+1))`

`=> P>=((a^2+b^2+c^2)^2)/(asqrt(b^2+1)+bsqrt(c^2+1)+csqrt(a^2+1))`

Mặt khác:

`[asqrt(b^2+1)+bsqrt(c^2+1)+csqrt(a^2+1)]^2<=(a^2+b^2+c^2)(b^2+1+c^2+1+a^2+1)`

`[asqrt(b^2+1)+bsqrt(c^2+1)+csqrt(a^2+1)]^2<=(a^2+b^2+c^2)(a^2+b^2+c^2+3)`

Mà `a^2+b^2+c^2>=ab+bc+ca=3`

`=> (a^2+b^2+c^2)(a^2+b^2+c^2+3)<=(a^2+b^2+c^2)(a^2+b^2+c^2+a^2+b^2+c^2)=2(a^2+b^2+c^2)^2`

`=> asqrt(b^2+1)+bsqrt(c^2+1)+csqrt(a^2+1)<=sqrt(2)(a^2+b^2+c^2)`

`=> P>=((a^2+b^2+c^2)^2)/(sqrt(2)(a^2+b^2+c^2))=(a^2+b^2+c^2)/(sqrt(2))`

`=> P>=(ab+bc+ca)/(sqrt(2))=3/(sqrt(2))=(3sqrt(2))/2`

Dấu "=" xảy ra khi `a=b=c=1`

Vậy GTNN của `P` là `(3sqrt(2))/2`