Cho a,b,c>0 thỏa mãn: c^2=a^3+b^3=3ab+c. Chứng minh rằng: 2a^2b^2+2b^2c^2+2c^2a^2=a^4+b^4+c^4.

Vì \({c^2} = 3ab + c\) nên \({c^3} = 3abc + {c^2}\)

Mà \({c^2} = {a^3} + {b^3}\) nên \({c^3} = 3abc + {a^3} + {b^3}\)

\({a^3} + {b^3} - {c^3} + 3abc = 0\)

\({\left( {a + b} \right)^3} - 3ab\left( {a + b} \right) - {c^3} + 3abc = 0\)

\({\left( {a + b} \right)^3} - {c^3} - 3ab\left( {a + b - c} \right) = 0\)

\(\left( {a + b - c} \right)\left[ {{{\left( {a + b} \right)}^2} + \left( {a + b} \right)c + {c^2}} \right] - 3ab\left( {a + b - c} \right) = 0\)

\(\left( {a + b - c} \right)\left( {{a^2} + {b^2} + {c^2} + ab + bc + ca} \right) = 0\)

Mà \(a,b,c > 0\) nên \({a^2} + {b^2} + {c^2} + ab + bc + ca > 0\)

Suy ra \(a + b - c = 0\)

Xét \(A = {a^4} + {b^4} + {c^4} - 2{a^2}{b^2} - 2{b^2}{c^2} - 2{c^2}{a^2}\)

\( = {\left( {{a^2} - {b^2}} \right)^2} + {c^4} - 2{b^2}{c^2} - 2{c^2}{a^2}\)

\( = {\left( {{a^2} - {b^2}} \right)^2} + {c^4} - 2\left( {{a^2} - {b^2}} \right){c^2} - 4{b^2}{c^2}\)

\( = {\left( {{a^2} - {b^2} - {c^2}} \right)^2} - {\left( {2bc} \right)^2}\)

\( = \left( {{a^2} - {b^2} - {c^2} - 2bc} \right)\left( {{a^2} - {b^2} - {c^2} + 2bc} \right)\)

\( = \left[ {{a^2} - {{\left( {b + c} \right)}^2}} \right]\left[ {{a^2} - {{\left( {b - c} \right)}^2}} \right]\)

\( = \left( {a - b - c} \right)\left( {a + b + c} \right)\left( {a - b + c} \right)\left( {a + b - c} \right)\)

Mà \(a + b - c = 0\)

Suy ra \(A = 0\)

Vậy \(2{a^2}{b^2} + 2{b^2}{c^2} + 2{c^2}{a^2} = {a^4} + {b^4} + {c^4}\).