bđt <=> x⁴ + y⁴ - x³y - xy³ ≥ 0

<=> x(x³ - y³) - y(x³- y³) ≥ 0

<=> x(x - y)(x² + xy + y²) - y(x - y)(x² + xy + y²) ≥ 0

<=> (x - y)²(x² + xy + y²) ≥ 0 (1)

Ta có: (x - y)² ≥ 0 ∀x, y

x² + xy + y² = (x + \(\dfrac{1}{2}\)y)² + \(\dfrac{3}{4}\)y² ≥ 0 ∀ x, y

=> (1) luôn đúng

Dấu "=" xảy ra <=> x = y

theo bđt cauchy schwars ta có:

\(\left\{{}\begin{matrix}x^4+y^4\ge2x^2y^2\\x^4+x^2y^2\ge2x^3y\\y^4+x^2y^2\ge2xy^3\end{matrix}\right.\)

\(\Leftrightarrow2\left(x^4+y^4\right)+2x^2y^2\ge2\left(xy^3+x^3y\right)+2x^2y^2\)

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

vậy đpcm

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

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

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

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

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

Ap dung bdt Cosi cho 2 so ko am la x² va y²/4 ta duoc

\(x^2+\frac{y^2}{4}\ge2\sqrt{\frac{x^2.y^2}{4}}=2.\frac{xy}{x}=xy\) 

Dau = xay ra <=> x=1/2y

Chuc  ban hoc tot

cai duoi mau la 2 nha chu ko phai la x dau , mik danh nhanh nen viet nham ban thong cam

a) \(\text{ }x^4+y^4\ge x^3y+xy^3\)

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

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

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

*NOTE: chứng minh đc vì (x-y)^2  >= 0 ;  x^2  +xy +y^2 > 0

mình cũng làm đến nơi rồi nhưng sợ x^2+xy+y^2 chưa chắc lớn hơn 0 thanks bạn nhé

ta có \(\left(x-y\right)^2\ge0\)

<=> \(x^2+y^2\ge2xy\)

<=>\(x^2+y^2+2xy\ge4xy\)

<=>\(\left(x+y\right)^2\ge4xy\)

<=>\(\frac{x+y}{xy}\ge\frac{4}{x+y}\)

<=>\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

\(xy\le\frac{\left(x+y\right)^2}{4}\)( bđt cauchy ) 

\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\)( bđt cauchy ) 

\(\Rightarrow\frac{x}{y}+\frac{y}{x}+\frac{xy}{\left(x+y\right)^2}\ge2+\frac{\frac{\left(x+y\right)^2}{4}}{\left(x+y\right)^2}=2+\frac{1}{4}=\frac{9}{4}\)

\(bdt< =>x\left(x+y\right)\le\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{y}< =>x^2-xy+y^2\ge xy\)

\(< =>\left(x-y\right)^2\ge0\)(dpcm)

Nhã Doanh giúp mk vs

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