Chứng minh: \({a^4} + {b^4} + {c^4} = 2{\left( {ab + bc + ac} \right)^2}\). Biết rằng \(a + b + c = 0\)
Giải chi tiết:
Ta có: \(a + b + c = 0 \Leftrightarrow a = - b - c.\)
\(\begin{array}{l} \Rightarrow {a^2} = {\left( {b + c} \right)^2} \Leftrightarrow {a^2} = {b^2} + {c^2} + 2bc\\ \Leftrightarrow {a^2} - {b^2} - {c^2} = 2bc.\\ \Leftrightarrow \left( {{a^2} - {b^2} - {c^2}} \right) = 4{b^2}{c^2}\\ \Leftrightarrow {a^4} + {b^4} + {c^4} - 2{a^2}{b^2} + 2{b^2}{c^2} - 2{a^2}{c^2} = 4{b^2}{c^2}\\ \Leftrightarrow {a^4} + {b^4} + {c^4} = 2{a^2}{b^2} + 2{b^2}{c^2} + 2{a^2}{c^2}\\ \Leftrightarrow {a^4} + {b^4} + {c^4} = 2\left( {{a^2}{b^2} + {b^2}{c^2} + {a^2}{c^2}} \right).\end{array}\)
Lại có:
\(\begin{array}{l}\;\;\;{\left( {ab + bc + ca} \right)^2} = {\left( {ab} \right)^2} + {\left( {bc} \right)^2} + {\left( {ca} \right)^2} + 2{a^2}bc + 2a{b^2}c + 2ab{c^2}\\ \Leftrightarrow \;{\left( {ab + bc + ca} \right)^2} = {a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + 2abc\left( {a + b + c} \right)\\ \Leftrightarrow \;{\left( {ab + bc + ca} \right)^2} = {a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}.\\ \Rightarrow {a^4} + {b^4} + {c^4} = 2{\left( {ab + bc + ca} \right)^2}.\;\;\;\left( {dpcm} \right)\end{array}\)