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a ) Ta có : \(a+b+c=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+ac+bc\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+ac+bc\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(a^2b^2+b^2c^2+c^2a^2+2ab^2c+2a^2bc+2c^2ab\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)+8abc\left(a+b+c\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)+8abc.0\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
Lại có : \(\dfrac{\left(a^2+b^2+c^2\right)^2}{2}=\dfrac{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}{2}\)
\(=\dfrac{a^4+b^4+c^4+a^4+b^4+c^4}{2}=\dfrac{2\left(a^4+b^4+c^4\right)}{2}\)
\(=a^4+b^4+c^4\left(đpcm\right)\)
b ) \(a+b+c+d=0\)
\(\Leftrightarrow a+b=-\left(c+d\right)\)
\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3+3a^2b+3b^2a+3c^2d+3d^2c=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(-a^2b-b^2a-c^2d-d^2c\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left[-ab\left(a+b\right)-cd\left(c+d\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left[ab\left(c+d\right)-cd\left(c+d\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\left(đpcm\right)\)
Ta có: \(a+b+c+d=a^2+b^2+c^2+d^2\)
\(\Rightarrow\orbr{\begin{cases}a=b=c=d=1\\a=b=c=d=0\end{cases}}\)
mà \(a^2+b^2+c^2+d^2=4\Rightarrow a=b=c=d=1\)
\(\Rightarrow ab+bc+cd+ad=1+1+1+1=4\)
Vậy.....
Ta có : \(\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)\ge\left(\sqrt{a^2b^2}+\sqrt{b^2c^2}+\sqrt{c^2d^2}\right)^2=\left(ab+bc+cd\right)^2\) (áp dụng bđt Schwartz)
Dấu " = " xảy ra khi \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Do đó, kết hợp cùng giả thiết suy ra đpcm