The least value of frac{x^2y^2 - 2x^2y + 2x^2 | vedclass

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(A) Let the expression be $f(x, y) = \frac{x^2y^2 - 2x^2y + 2x^2 + 2xy - 2x + 1}{x^2y + x}$.We can rewrite the numerator as $(xy + 1)^2 - 2x(xy + 1) +

Vedclass · Accepted Answer

$\lambda \in (0, 1)$

Vedclass · Answer

(A) Let the expression be $f(x, y) = \frac{x^2y^2 - 2x^2y + 2x^2 + 2xy - 2x + 1}{x^2y + x}$.We can rewrite the numerator as $(xy + 1)^2 - 2x(xy + 1) + 2x^2$.Thus,$f(x, y) = \frac{(xy + 1)^2 - 2x(xy + 1) + 2x^2}{x(xy + 1)}$.Dividing each term by $x(xy + 1)$,we get $f(x, y) = \frac{xy + 1}{x} - 2 + \frac{2x}{xy + 1}$.Let $u = \frac{xy + 1}{x}$. Since $x, y > 0$,$u = y + \frac{1}{x} > 0$.Then $f(x, y) = u + \frac{2}{u} - 2$.By the Arithmetic Mean-Geometric Mean Inequality $(AM \geq GM)$,$u + \frac{2}{u} \geq 2\sqrt{u \cdot \frac{2}{u}} = 2\sqrt{2}$.Therefore,the minimum value $\lambda = 2\sqrt{2} - 2$.Since $\sqrt{2} \approx 1.414$,$\lambda \approx 2(1.414) - 2 = 2.828 - 2 = 0.828$.Thus,$0 < \lambda < 1$,which means $\lambda \in (0, 1)$.

The least value of $\frac{x^2y^2 - 2x^2y + 2x^2 + 2xy - 2x + 1}{x^2y + x}$ is $\lambda$,where $x, y \in R^+$ and $x^2y + x \neq 0$. Then:

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