Question

Answer 1

-Áp dụng BĐT AM-GM ta có:

\(\left(a^2+1\right)\left(a^2b^2+4\right)\left(a^2b^2c^2+16\right)\ge2a.4ab.8abc=64a^3b^2c\)

Answer 2

Áp dụng BDDT Cauchy:

\(a^2+1\ge2a\)

\(a^2b^2+4\ge2\sqrt{a^2b^2.4}=4ab\)

\(a^2b^2c^2+16\ge2\sqrt{a^2b^2c^2.16}=8abc\)

Nhân vế với vế:

\(\left(a^2+1\right)\left(a^2b^2+4\right)\left(a^2b^2c^2+16\right)\ge64a^3b^2c\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;2;2\right)\)