Need help in solving an inequality

Question

The inequality $px^2 + qxy + ry^2 > 0$ should hold true for all possible real values of $x$ and $y$ (both $x$ and $y$ not equal to zero).
How do I find the values of $p$, $q$ and $r$ such that the above condition holds true ?

Answer 1

Hint: put $y=0,x=1$ to see that $p>0$. Similarly $r>0$. Now write the in equality as $p(x+\frac q {2p}y)^{2}+\frac {4pr-q^{2}} {4p}y^{2}$. Now conclude that the necessary and sufficient conditions are $p>0, r>0$ and $q^{2}<4pr$.

Answer 2

If you know already something about linear algebra and quadratic forms you can proceed as follows:

• $px^2 + qxy + ry^2 = (x\; y)\underbrace{\begin{pmatrix}p & \frac{q}{2}\\\frac{q}{2} & r \end{pmatrix}}_{A:=}\begin{pmatrix}x \\y \end{pmatrix}$

Your condition means that $A$ should be positive definite which is equivalent to $$p > 0 \mbox{ and } \det (A) = pr-\frac{q^2}{4} > 0$$.