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

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

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

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

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

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

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

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

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

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

:))

\(x^4+y^4\ge xy^3+x^3y\\ \Leftrightarrow x^4-xy^3+y^4-x^3y\ge0\\ \Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\\ \Leftrightarrow\left(x-y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\ge0\\ \Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\\ \)

Ma \(\left(x-y\right)^2\ge0\left(\forall x,y\right)\\ x^2+xy+y^2>0\left(\forall x,y\right)\)\(\Rightarrow x^4+y^4\ge xy^3+x^3y\left(\forall x,y\right)\left(dpcm\right)\)

Đề là CMR $x^4-x^3y+x^2y^2-xy^3+y^4> x^2+y^2$ thì đúng hơn bạn ạ.

Lời giải:

Ta có:

$\text{VT}=(x^4+y^4-x^3y-xy^3)+x^2y^2$

$=(x-y)^2(x^2+xy+y^2)+x^2y^2\geq x^2y^2$

Mà:

$x^2y^2=\frac{x^2y^2}{2}+\frac{x^2y^2}{2}> \frac{x^2.2}{2}+\frac{2.y^2}{2}=x^2+y^2$ do $x^2> 2, y^2>2$

Do đó: $\text{VT}> x^2+y^2$ (đpcm)

đề đúng \(x^4+y^4\ge xy^3+x^3y\)

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

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

Dấu = khi x=y

"=" là dấu "+" đó các bạn

Ta có (x¹⁰ - y¹⁰) = (x⁵ + y⁵)(x⁵ - y⁵) = (x⁵ - y⁵)(x + y)(x⁴ - x³ y + x² y² - xy³ + y⁴)

Xong

\(2\left(x^4+y^4\right)\ge xy^3+x^3y+2x^2y^2\)

\(\Leftrightarrow\left(x^4-2x^2y^2+y^4\right)+\left(x^4-x^3y\right)+\left(y^4-xy^3\right)\ge0\)

\(\Leftrightarrow\left(x^2-y^2\right)^2+x^3\left(x-y\right)+y^3\left(y-x\right)\ge0\)

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

\(\Leftrightarrow\left(x^2-y^2\right)^2+\left(x-y\right)^2\left[\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{2}\right]\ge0\) ( đúng )