Áp dụng bđt sau \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)
Có: \(A=\left(x+2\right)^4+\left(x-4\right)^4\)
\(=\left(x+2\right)^4+\left(4-x\right)^4\)
\(\ge\frac{\left[\left(x+2\right)^2+\left(4-x\right)^2\right]^2}{2}\ge\frac{\left[\frac{\left(x+2+4-x\right)^2}{2}\right]^2}{2}\)
\(=\frac{\left(\frac{6^2}{2}\right)^2}{2}=162\)
Dấu "=" xảy ra <=> x + 2 = 4 - x
<=> 2x = 2
<=> x = 1