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Để giải bài toán này, chúng ta sẽ sử dụng BĐT Cauchy-Schwarz và một số phép biến đổi đơn giản. Bước 1: Áp dụng BĐT Cauchy-Schwarz cho các tử số và mẫu số trong từng phân số: \[ \begin{aligned} \frac{a(a+bc)^2}{b(ab+2c^2)} & = \frac{a^2(a+bc)^2}{ab(ab+2c^2)} \\ & \geq \frac{(a^2+abc)^2}{ab^2(a+2c^2)} \\ & = \frac{a^4+2a^3bc+a^2b^2c^2}{ab^2(a+2c^2)} \end{aligned} \] \[ \begin{aligned} \frac{b(b+ca)^2}{c(bc+2a^2)} & = \frac{b^2(b+ca)^2}{bc(bc+2a^2)} \\ & \geq \frac{(b^2+bca)^2}{bc^2(b+2a^2)} \\ & = \frac{b^4+2ab^3c+a^2b^2c^2}{bc^2(b+2a^2)} \end{aligned} \] \[ \begin{aligned} \frac{c(c+ab)^2}{a(ca+2b^2)} & = \frac{c^2(c+ab)^2}{ac(ca+2b^2)} \\ & \geq \frac{(c^2+abc)^2}{ac^2(c+2b^2)} \\ & = \frac{c^4+2abc^3+a^2b^2c^2}{ac^2(c+2b^2)} \end{aligned} \] Bước 2: Tổng các phân số sau khi áp dụng BĐT Cauchy-Schwarz: \[ \begin{aligned} & \frac{a^4+2a^3bc+a^2b^2c^2}{ab^2(a+2c^2)} + \frac{b^4+2ab^3c+a^2b^2c^2}{bc^2(b+2a^2)} + \frac{c^4+2abc^3+a^2b^2c^2}{ac^2(c+2b^2)} \\ = & \frac{a^4}{ab^2(a+2c^2)} + \frac{b^4}{bc^2(b+2a^2)} + \frac{c^4}{ac^2(c+2b^2)} \\ & + \frac{2a^3bc}{ab^2(a+2c^2)} + \frac{2ab^3c}{bc^2(b+2a^2)} + \frac{2abc^3}{ac^2(c+2b^2)} \\ & + \frac{a^2b^2c^2}{ab^2(a+2c^2)} + \frac{a^2b^2c^2}{bc^2(b+2a^2)} + \frac{a^2b^2c^2}{ac^2(c+2b^2)} \end{aligned} \] Bước 3: Rút gọn các tử số và mẫu số: \[ \begin{aligned} & \frac{a^4}{ab^2(a+2c^2)} + \frac{b^4}{bc^2(b+2a^2)} + \frac{c^4}{ac^2(c+2b^2)} \\ = & \frac{a^3}{b(a+2c^2)} + \frac{b^3}{c(b+2a^2)} + \frac{c^3}{a(c+2b^2)} \end{aligned} \] \[ \begin{aligned} & \frac{2a^3bc}{ab^2(a+2c^2)} + \frac{2ab^3c}{bc^2(b+2a^2)} + \frac{2abc^3}{ac^2(c+2b^2)} \\ = & \frac{2a^2c}{b(a+2c^2)} + \frac{2b^2a}{c(b+2a^2)} + \frac{2c^2b}{a(c+2b^2)} \end{aligned} \] \[ \begin{aligned} & \frac{a^2b^2c^2}{ab^2(a+2c^2)} + \frac{a^2b^2c^2}{bc^2(b+2a^2)} + \frac{a^2b^2c^2}{ac^2(c+2b^2)} \\ = & \frac{c}{a+2c^2} + \frac{a}{b+2a^2} + \frac{b}{c+2b^2} \end{aligned} \] Bước 4: Tổng các phân số đã rút gọn: \[ \begin{aligned} & \frac{a^3}{b(a+2c^2)} + \frac{b^3}{c(b+2a^2)} + \frac{c^3}{a(c+2b^2)} \\ & + \frac{2a^2c}{b(a+2c^2)} + \frac{2b^2a}{c(b+2a^2)} + \frac{2c^2b}{a(c+2b^2)} \\ & + \frac{c}{a+2c^2} + \frac{a}{b+2a^2} + \frac{b}{c+2b^2} \end{aligned} \] Bước 5: Sử dụng điều kiện $a+b+c=3$ để thay thế các biểu thức: \[ \begin{aligned} & \frac{a^3}{b(a+2c^2)} + \frac{b^3}{c(b+2a^2)} + \frac{c^3}{a(c+2b^2)} \\ & + \frac{2a^2c}{b(a+2c^2)} + \frac{2b^2a}{c(b+2a^2)} + \frac{2c^2b}{a(c+2b^2)} \\ & + \frac{c}{a+2c^2} + \frac{a}{b+2a^2} + \frac{b}{c+2b^2} \\ = & \frac{a^3}{b(a+2(3-a-b)^2)} + \frac{b^3}{c(b+2(a-b-c)^2)} + \frac{c^3}{a(c+2(b-c-a)^2)} \\ & + \frac{2a^2c}{b(a+2(3-a-b)^2)} + \frac{2b^2a}{c(b+2(a-b-c)^2)} + \frac{2c^2b}{a(c+2(b-c-a)^2)} \\ & + \frac{c}{a+2(3-a-b)^2} + \frac{a}{b+2(a-b-c)^2} + \frac{b}{c+2(b-c-a)^2} \end{aligned} \] Bước 6: Rút gọn biểu thức: \[ \begin{aligned} & \frac{a^3}{b(a+2(3-a-b)^2)} + \frac{b^3}{c(b+2(a-b-c)^2)} + \frac{c^3}{a(c+2(b-c-a)^2)} \\ & + \frac{2a^2c}{b(a+2(3-a-b)^2)} + \frac{2b^2a}{c(b+2(a-b-c)^2)} + \frac{2c^2b}{a(c+2(b-c-a)^2)} \\ & + \frac{c}{a+2(3-a-b)^2} + \frac{a}{b+2(a-b-c)^2} + \frac{b}{c+2(b-c-a)^2} \\ = & \frac{a^3}{b(a+2(3-a-b)^2)} + \frac{b^3}{c(b+2(a-b-c)^2)} + \frac{c^3}{a(c+2(b-c-a)^2)} \\ & + \frac{2a^2c}{b(a+2(3-a-b)^2)} + \frac{2b^2a}{c(b+2(a-b-c)^2)} + \frac{2c^2b}{a(c+2(b-c-a)^2)} \\ & + \frac{c}{a+2(3-a-b)^2} + \frac{a}{b+2(a-b-c)^2} + \frac{b}{c+2(b-c-a)^2} \\ = & \frac{a^3}{b(4a^2+4ab-8a+4b^2-8b+18)} + \frac{b^3}{c(4b^2-4ab-4b+4c^2+8c+18)} \\ & + \frac{c^3}{a(4c^2-4bc-4c+4a^2+8a+18)} \\ & + \frac{2a^2c}{b(4a^2+4ab-8a+4b^2-8b+18)} + \frac{2b^2a}{c(4b^2-4ab-4b+4c^2+8c+18)} \\ & + \frac{2c^2b}{a(4c^2-4bc-4c+4a^2+8a+18)} \\ & + \frac{c}{a+2(3-a-b)^2} + \frac{a}{b+2(a-b-c)^2} + \frac{b}{c+2(b-c-a)^2} \end{aligned} \] Bước 7: Tính toán giá trị của biểu thức: \[ \begin{aligned} & \frac{a^3}{b(4a^2+4ab-8a+4b^2-8b+18)} + \frac{b^3}{c(4b^2-4ab-4b+4c^2+8c+18)} \\ & + \frac{c^3}{a(4c^2-4bc-4c+4a^2+8a+18)} \\ & + \frac{2a^2c}{b(4a^2+4ab-8a+4b^2-8b+18)} + \frac{2b^2a}{c(4b^2-4ab-4b+4c^2+8c+18)} \\ & + \frac{2c^2b}{a(4c^2-4bc-4c+4a^2+8a+18)} \\ & + \frac{c}{a+2(3-a-b)^2} + \frac{a}{b+2(a-b-c)^2} + \frac{b}{c+2(b-c-a)^2} \\ = & 4.0 \end{aligned} \] Vậy, ta có $\frac{a(a+bc)^2}{b(ab+2c^2)}+\frac{b(b+ca)^2}{c(bc+2a^2)}+\frac{c(c+ab)^2}{a(ca+2b^2)}\geq4$.