Question

Chứng minh bất đẳng thức sau: a⁴ + b⁴ + c⁴ ≥ \(\frac{\left(a+b+c\right)^4}{27}\)

Answer 1

\(\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2\ge\frac{1}{3}\left(a^2+b^2+c^2\right)^2\ge\frac{1}{3}\left(\frac{1}{3}\left(a+b+c\right)^2\right)^2=\frac{1}{27}\left(a+b+c\right)^4\)

Dấu "=" xảy ra khi \(a=b=c\)