a)AM-GM:
\(a^4+a^4+b^4+c^4\ge4\sqrt[4]{a^4\cdot a^4\cdot b^4\cdot c^4}=4a^2bc\)
\(a^4+b^4+b^4+c^4\ge4ab^2c\)
\(a^4+b^4+c^4+c^4\ge4abc^2\)
Cộng vế theo vế ta được:
4\(\left(a^4+b^4+c^4+d^4\right)\ge4a^2bc+4ab^2c+4abc^2\)
\(\Leftrightarrow a^4+b^4+c^4+d^4\ge abc\left(a+b+c\right)\)
1 cách khác: \(a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)
\(2\left(a^2b^2+b^2c^2+a^2c^2\right)\ge2\sqrt{a^2b^4c^2}+2\sqrt{b^2a^2c^4}+2\sqrt{a^4b^2c^2}\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2\ge ab^2c+abc^2+a^2bc=abc\left(a+b+c\right)\)
\(\Rightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
tương tự với câu b
b)AM-GM:
\(a^8+a^8+b^8+c^8\ge4a^4b^2c^2\)
\(a^8+b^8+b^8+c^8\ge4a^2b^4c^2\)
\(a^8+b^8+c^8+c^8\ge4a^2b^2c^4\)
Cộng vế theo vế ta được:
\(4\left(a^8+b^8+c^8\right)\ge4\left(a^4b^2c^2+a^2b^4c^2+a^2b^2c^4\right)\)
\(\Leftrightarrow a^8+b^8+c^8\ge a^2b^2c^2\left(a^2+b^2+c^2\right)\ge a^2b^2c^2\left(ab+bc+ca\right)\left(đpcm\right)\)