Prove that : A = x2 - xy + y2 > 0 ( ; )

Question by Kaya Renger

13/09/2017 at 12:33

A = x² - xy + y² > 0 ($\forall x,y$ ; $A\ne0$)

Dao Trong Luan 13/09/2017 at 13:11

$A=x^2-xy+y^2>x^2-xy-xy+y^2$

$\Rightarrow A>x^2-2xy+y^2$

$\Rightarrow A>\left(x-y\right)^2$

But $\left(x-y\right)^2\ge0$

$\Leftrightarrow A\ge0$

But $A\ne0\Rightarrow A>0$
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Vũ Trung Dũng 30/09/2017 at 18:07

$A = x^{2} - x y + y^{2} > x^{2} - x y - x y + y^{2}$

$\Rightarrow A > x^{2} - 2 x y + y^{2}$

$\Rightarrow A > {(x - y)}^{2}$

But ${(x - y)}^{2} \geq 0$

$\Leftrightarrow A \geq 0$

But $A \neq 0 \Rightarrow A > 0$
Dinh Hung 28/09/2017 at 21:29

We have : $A=x^2-xy+y^2=\left(x^2-xy+\dfrac{1}{4}y^2\right)+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x;y;A\ne0$