Harder Inequalities question from Cambridge Book Show $x^2+xy+y^2>0$ for $x, y\ne 0$.

Harder Inequalities question from Cambridge Book Show $$x^2+xy+y^2>0$$ for $x, y\ne 0$.

3 Methods:

Method 1: Consider two cases. (Keep in mind $x,y$ are both non zero)
Case 1: $xy>0$. In this case the result is trivial as each summand is positive. That is since $x^2>0, y^2>0$ and $xy>0$ then their sum is also positive!//

Case 2: $xy<0$. Then $-xy>0$.
So $x^2+xy+y^2=(x+y)^2-xy$. But $(x+y)^2>0$ and so the LHS is the sum of two positive numbers and so is itself positive.//

Method 2: Use completing the square

$\begin{array}{l} x^2+xy+y^2&=x^2+xy+y^2/4+3y^2/4\\ &=\left( x+y/2 \right)^2 +3y^2/4\end{array}$

Each summand is clearly positive and hence the sum is also.//

Method 3: Use the discriminant by considering $x^2+xy+y^2$ as a quadratic in $x$.
$\Delta = y^2-4y^2=-3y^2<0$.
Since the coefficient of $x$ is $1>0$ and $\Delta<0$ then the expression must be positive.

[Remember that if you have a quadratic $Ax^2+Bx+C$ then $\Delta=B^2-4AC$]