`A=\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}`
`=\sqrt{a^2+2ab+b^2-ab}+\sqrt{b^2+2bc+c^2-bc}+\sqrt{c^2+2ca+a^2-ca}`
`=\sqrt{(a+b)^2-ab}+\sqrt{(b+c)^2-bc}+\sqrt{(c+a)^2-ca}`
Lại có: `(a-b)^2>=0`
`<=>a^2-2ab+b^2>=0`
`<=>a^2+2ab+b^2>=4ab`
`<=>(a+b)^2>=4ab`
`<=>(a+b)^2/4>=ab`
`<=>-ab>=-(a+b)^2/4`
Hoàn toàn tương tự:
`-bc>=-(b+c)^2/4;\ -ca>=-(c+a)^2/4`
`=>A>=\sqrt{(a+b)^2-(a+b)^2/4}+\sqrt{(b+c)^2-(b+c)^2/4}+\sqrt{(c+a)^2-(c+a)^2/4}`
`=\sqrt{{3(a+b)^2}/4}+\sqrt{{3(b+c)^2}/4}+\sqrt{{3(c+a)^2}/4}`
`={\sqrt{3}(a+b)}/2+{\sqrt{3}(b+c)}/2+{\sqrt{3}(c+a)}/2`
`=\sqrt3/2(a+b+b+c+c+a)=\sqrt3/2 .2(a+b+c)`
`=\sqrt3 . \sqrt3 =3`
Dấu `=` xảy ra `<=>a=b=c=\sqrt3/3`
Vậy `A_{min}=3` đạt được khi `a=b=c=\sqrt3/3`
