Question

Cho a, b là 2 số dương. Chứng minh: \(a^4+b^4\ge\frac{\left(a+b\right)^4}{8}\)

Answer 1

qua de

\(a^4+b^4\ge\frac{\left(a^2+b^2\right)^2}{2}\ge\frac{\left(a+b\right)^4}{8}\)

Answer 2

áp dụng BĐT bnyacovsky :\(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow2a^2+2b^2\ge\left(a+b\right)^2\)

\(\left(4+4\right)\left(a^4+b^4\right)\ge\left(2a^2+2b^2\right)^2\ge\left(a+b\right)^4\)

\(\Leftrightarrow a^4+b^4\ge\frac{\left(a+b\right)^4}{8}\)

dấu = xảy ra khi a=b