[tex] {x}^{2} + {y}^{2} \geqslant {x}^{2} y + x {y}^{2} [/tex]

Предыдущий
вопрос Следующий
вопрос

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 $x$ (assuming $x \neq 0$), 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 $x^2 + y^2 \geq x^2y + xy^2$ is the same as the fact that $x \geq y(x+y)$ for $x,y \in \mathbb{R}$.