\(a^4-\left(b^2-c^2\right)^2=a^4-\left(b-c\right)^2\left(b+c\right)^2=a^4-a^2\left(b-c\right)^2\)
\(=a^2\left[a^2-\left(b-c\right)^2\right]=a^2\left(a-b+c\right)\left(a+b-c\right)=a^2.\left(-2b\right).\left(-2c\right)=4a^2bc\)
Tương tự:
\(b^4-\left(c^2-a^2\right)^2=4ab^2c\)
\(c^4-\left(a^2-b^2\right)^2=4abc^2\)
Mặt khác với \(a+b+c=0\) ta có:
\(a+b=-c\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=-3ab.\left(-c\right)=3abc\)
Đặt vế trái là P:
\(\Rightarrow P=\dfrac{a^4}{4a^2bc}+\dfrac{b^4}{4ab^2c}+\dfrac{c^4}{4abc^2}=\dfrac{1}{4}\left(\dfrac{a^3+b^3+c^3}{abc}\right)=\dfrac{1}{4}.\dfrac{3abc}{abc}=\dfrac{3}{4}\)
