Ta có: \(a+b+c=0\)
\(\Rightarrow\)\((a+b+c)^2=0\)
\(\Rightarrow\)\(a^2+b^2+c^2+2ab+2ac+2bc=0\)
\(\Rightarrow\)\(1+2(ab+bc+ac)=0\) ( Vì \(a^2+b^2+c^2=1\) )
\(\Rightarrow\)\(ab+bc+cd=\)\(-\dfrac{1}{2}\)
\(\Rightarrow\)\((ab+bc+cd)^2=\)\(\dfrac{1}{4}\)
\(\Rightarrow\)\(a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2\)\(=\)\(\dfrac{1}{4}\)
\(\Rightarrow\)\(a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c)\)\(=\dfrac{1}{4}\)
\(\Rightarrow\)\(a^2b^2 +a^2c^2+b^2c^2\)\(=\dfrac{1}{4}\) ( Vì \(a+b+c=0 \)) \((1)\)
Mặt khác: \(a^2+b^2+c^2=1\)
\(\Rightarrow\)\((a^2+b^2+c^2)^2=1\)
\(\Rightarrow\)\(a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1\)
\(\Rightarrow\)\(a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)=1\)
\(\Rightarrow\)\(a^4+b^4+c^4+2.\)\(\dfrac{1}{4}=1\) (Theo \(1\))
\(\Rightarrow\)\(a^4+b^4+c^4 \)\(=1-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\Rightarrow\) Đpcm.
