Dividing both sides by xxx (assuming x≠0x \neq 0x=0), we get:
[x(1-y) \geq y^2]
[x - xy \geq y^2]
[x \geq y^2 + xy]
[x \geq y(x+y)]
Now, the fact that x2+y2≥x2y+xy2x^2 + y^2 \geq x^2y + xy^2x2+y2≥x2y+xy2 is the same as the fact that x≥y(x+y)x \geq y(x+y)x≥y(x+y) for x,y∈Rx,y \in \mathbb{R}x,y∈R.
The given inequality is equivalent to:
[x^2 + y^2 - x^2y - xy^2 \geq 0]
[x^2(1-y) - y^2x \geq 0]
[x^2(1-y) \geq y^2x]
Dividing both sides by xxx (assuming x≠0x \neq 0x=0), we get:
[x(1-y) \geq y^2]
[x - xy \geq y^2]
[x \geq y^2 + xy]
[x \geq y(x+y)]
Now, the fact that x2+y2≥x2y+xy2x^2 + y^2 \geq x^2y + xy^2x2+y2≥x2y+xy2 is the same as the fact that x≥y(x+y)x \geq y(x+y)x≥y(x+y) for x,y∈Rx,y \in \mathbb{R}x,y∈R.