`1/(b + c)^2 + 1/(c + a)^2 + 1/(a + b)^2 >= (3sqrt(3abc(a + b + c))(a + b + c)^2)/(4(ab + bc + ca)^3)`

Đặt `1/x = a ; 1/y = b ; 1/z = c`

Bất đẳng thức cần chứng minh trở thành `:`

`1/(1/y + 1/z)^2 + 1/(1/z + 1/x)^2 + 1/(1/x + 1/y)^2`

`>= (3 sqrt(3/(xyz) . (1/x + 1/y + 1/z))(1/x + 1/y + 1/z)^2)/(4(1/(xy) + 1/(yz) + 1/(zx))^3)`

`<=> 1/(1/y^2 + 2/(yz) + 1/z^2) + 1/(1/z^2 + 2/(zx) + 1/x^2) + 1/(1/x^2 + 2/(xy) + 1/y^2)`

`>=  (3. sqrt(3(xy + yz + zx)) . 1/(xyz) . ((xy + yz + zx)/(xyz))^2)/(4((x + y + z)/(xyz))^3`

`<=>  (y^2z^2)/(y + z)^2 + (z^2x^2)/(x + z)^2 + (x^2y^2)/(x + y)^2`

`>= (3sqrt(3(xy + yz+ zx)) (xy + yz + zx)^2)/(4(x + y + z)^3`

Áp dụng bất đẳng thức Cauchy Schwarz : 

`  (y^2z^2)/(y + z)^2 + (z^2x^2)/(x + z)^2 + (x^2y^2)/(x + y)^2`

`>= (xy + yz + zx)^2/((x + y)^2 + (y + z)^2 + (z + x)^2)`

Ta cần chứng minh `:`

` (xy + yz + zx)^2/((x + y)^2 + (y + z)^2 + (z + x)^2) `

`>= (3sqrt(3(xy + yz+ zx)) (xy + yz + zx)^2)/(4(x + y + z)^3`

`<=>  4(xy + yz + zx)^2(x + y + z)^3`

`>= 3sqrt(3(xy + yz+ zx)) (xy + yz + zx)^2 . [(x + y)^2 + (y + z)^2 + (z + x)^2]`

`<=> 4(x + y + z)^3 `

`>= 3sqrt(3(xy + yz+ zx)) . [(x + y)^2 + (y + z)^2 + (z + x)^2]`

`<=> 4(x + y + z)^3`

`>= 3sqrt(3(xy + yz+ zx)) . [2(x^2 + y^2 + z^2) + 2(xy + yz + zx)]`

`<=> 2(x + y + z)^3`

`>= 3sqrt(3(xy + yz+ zx)) (x^2 + y^2 + z^2 + xy + yz + zx)`

`<=> 2(x + y + z)^2 . (x + y + z)`

`>= 3sqrt(3(xy + yz+ zx)) (x^2 + y^2 + z^2 + xy + yz + zx)`

`<=> 2[x^2 + y^2 + z^2 + 2(xy + yz + zx)] . sqrt((x + y + z)^2) >= 3sqrt(3(xy + yz+ zx)) (x^2 + y^2 + z^2 + xy + yz + zx)`  `(1)`

Đặt `m = x^2 + y^2 + z^2 ; n = xy + yz + zx`  `(m;m > 0)`

`(1) <=> 2(m + 2n)sqrt(m + 2n) >= 3(m + n)sqrt(3n)`

`<=> 4(m + 2n)^2 (m + 2n) >= 9(m + n)^2 . 3n`

`<=> 4(m + 2n)^3 >= 27n(m + n)^2`

`<=> 4(m + 2n)^3 - 27n(m + n)^2 >= 0`

`<=> 4 [m^3 + 8n^3 + 6mn(m+2n)] - 27n(m^2 + 2mn+ n^2 ) >= 0`

`<=> 4(m^3 + 8n^3 + 6m^2n + 12mn^2) - 27mn^2 - 54mn^2 - 27n^3 >= 0`

`<=> 4m^3 + 32n^3 + 24m^2 n + 48mn^2 - 27mn^2 - 54mn^2 - 27n^3 >= 0`

`<=> 4m^3 + 5n^3 - 3m^2n - 6mn^2 >= 0`

`<=> 4m^3 -4 m^2 n + m^2 n - mn^2 - 5mn^2 + 5mn^3 >= 0`

`<=> 4m^2 (m-n) + mn(m-n) - 5n^2 (m-n) >=  0`

`<=> (m-n)(4m^2 + mn - 5n^2) >=  0`

`<=> (m-n)(4m^2 - 4mn + 5mn - 5n^2) >=  0`

`<=> ( m - n)[4m(m - n) + 5n( m - n)] >= 0`

`<=> (m - 2)(m - 2)(4m + 5m) >= 0`

`<=> (m-n)^2 (4m + 5n) >= 0` (Luôn đúng)

Vậy BĐT được chứng minh

Dấu "=" xảy ra khi `a=b=c`