a+b+c=0
\(\Leftrightarrow\)(a+b+c)2=0
\(\Leftrightarrow\)a2+b2+c2+2ab+2bc+2ca=0 mà a2+b2+c2=2
\(\Rightarrow\)2ab+2bc+2ca=-2
\(\Leftrightarrow\)(2ab+2bc+2c)2=4
\(\Leftrightarrow\)4a2b2+4c2b2+4a2c2+8abc(a+b+c)=4 mà a+b+c=0
\(\Rightarrow\)4a2b2+4c2b2+4a2c2=4 (1)
\(\Leftrightarrow\)2a2b2+2c2b2+2a2c2=2
Mặt khác:
a2+b2+c2=2 \(\Rightarrow\)(a2+b2+c2)2=4
\(\Leftrightarrow\)a4+b4+c4+2(a2b2+b2c2+c2a2)=4 (2)
Từ (1) và (2) \(\Rightarrow\)4a2b2+4c2b2+4a2c2=a4+b4+c4+2(a2b2+b2c2+c2a2)
\(\Leftrightarrow\)2a2b2+2c2b2+2a2c2=a4+b4+c4
\(\Rightarrow\)a4+b4+c4=2 (vì 2a2b2+2c2b2+2a2c2=2)
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