Question

cmr x^4+y^4 >=xy^3+x^3y với mọi x,y

Answer 1

\(x^4+y^4\ge xy^3+x^3y\)

\(x^4+y^4-xy^3-x^3y\ge0\)

\(x^3\left(x-y\right)+y^3\left(y-x\right)\ge0\)

\(x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)

\(\left(x-y\right)\left(x^3-y^3\right)\ge0\)

\(\left(x-y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\ge0\)

\(\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)

Dễ dàng c/m được \(x^2+y^2>xy\Rightarrow\left(x^2+xy+y^2\right)\ge0\)

\(\Rightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\left(đpcm\right)\)

Answer 2

Cái chỗ \(x^2+y^2>xy\)phải là \(x^2+y^2\ge0\)nha -_-"

Dấu "=" <=> x=y=0

Chứng minh nè :

\(x^2+y^2=\left(x+y\right)^2-2xy\ge2xy\ge xy\)

Dấu "=" xảy ra <=> x=y=0

:))