`a;b;c>0`
Áp dụng BĐT Cosi cho `3` số dương ta có:
`\sqrt{{a^2+2b^2}/{a^2+ab+bc}}+\sqrt{{b^2+2c^2}/{b^2+bc+ca}}+\sqrt{{c^2+2a^2}/{c^2+ca+ab}}\ge `$3\sqrt[3]{\sqrt{\dfrac{(a^2+2b^2)(b^2+2c^2)(c^2+2a^2)}{(a^2+ab+bc)(b^2+bc+ca)(c^2+ca+ab)}}}\quad (1)$
BĐT Bunhiacopxki với bộ số thực `(a_1;a_2;a_3); (b_1;b_2;b_3)`:
`\quad (a_1^2+a_2^2+a_3^2).(b_1^2+b_2^2+b_3^2)`
`\ge (a_1b_1+a_2b_2+a_3b_3)^2`
Dấu "=" xảy ra khi: `{a_1}/{b_1}={a_2}/{b_2}={a_3}/{b_3}`
Áp dụng BĐT Bunhiacopxki ta có:
`a^2+2b^2=b^2+a^2+b^2`
`b^2+2c^2=b^2+c^2+c^2`
`=>(a^2+2b^2)(b^2+2c^2)`
`=(b^2+a^2+b^2).(b^2+c^2+c^2)`
`\ge (b.b+a.c+b.c)^2`
`=>(a^2+2b^2)(b^2+2c^2)\ge (b^2+ac+bc)^2`
Tương tự chứng minh được:
`(a^2+2b^2)(c^2+2a^2)\ge (a^2+bc+ab)^2`
`(b^2+2c^2)(c^2+2a^2)\ge (c^2+ab+ac)^2`
`=>(a^2+2b^2)^2.(b^2+2c^2)^2 .(c^2+2a^2)^2`
`\ge (b^2+ac+bc)^2 . (a^2+bc+ab)^2 .(c^2+ab+ac)^2`
`=>(a^2+2b^2) .(b^2+2c^2) .(c^2+2a^2)`
`\ge (b^2+ac+bc) . (a^2+bc+ab).(c^2+ab+ac)`
`=>(a^2+2b^2) .(b^2+2c^2) .(c^2+2a^2)`
`\ge (a^2+ab+bc) . (b^2+bc+ac).(c^2+ca+ab)`
$\\$
`=>{(a^2+2b^2) .(b^2+2c^2) .(c^2+2a^2)}/{(a^2+ab+bc) . (b^2+bc+ac).(c^2+ca+ab)}>1\ (2)` (vì hai vế đều `>0)`
Từ `(1);(2)`
`=>\sqrt{{a^2+2b^2}/{a^2+ab+bc}}+\sqrt{{b^2+2c^2}/{b^2+bc+ca}}+\sqrt{{c^2+2a^2}/{c^2+ca+ab}}\ge 3`
Dấu "=" xảy ra khi `a=b=c`
