Contest Math Linear Algebra problem

Question

If $A\in M_2(\mathbb{R})$, prove that: $$ \det(A^2 + A + I_2) \geq \frac{3}{4}(1-\det A)^2.$$

If $x,y$ are the eigenvalues of $A$ then we get the following inequality $$(x^2+x+1)(y^2+y+1)\geq \frac{3}{4}(1-xy)^2.$$ I was stuck after this step and it turns out that $$ (x ^2 + x +1)(y^2 +y +1) - \frac{3}{4} \left( xy-1 \right)^2 = \frac{1}{4}(2x +2 y +xy +1 )^2 \geq 0.$$ My question is how does one go about finding such a complete square? Are there some tricks that can help one to find such expressions? Perhaps someone knows a condition under which the equation of the a general conic

$$ax^2+hxy+by^2+cx+dy+f=0$$ is a perfect square?

Answer 1

I would try like this. The following equation must be true for all $x,y$:

$$ (x ^2 + x +1)(y^2 +y +1) - \frac{3}{4} \left( xy-1 \right)^2 = (ax^2+hxy+by^2+cx+dy+f)^2$$

So what would we get if $y=0$: $$ x ^2 + x +1 - \frac{3}{4} = (ax^2+cx+f)^2$$

so $$ (x+{1\over 2})^2 = (ax^2+cx+f)^2$$ Since this holds for all $x$ we have $a=0$, $c= 1$ and $f = {1\over 2}$.

With the same procedure for $x=0$ we get $b=0$ and $d=1$. We are left to calculate $h$.

For $x=y=-1$ we get then $$(h+{5\over 2})^2=9$$ and for $x=y=1$ we get then $$(h-{3\over 2})^2=1$$ so $h={1\over 2}$.

Answer 2

I can't think of a better way than completing square. Anyway, as the matrices are $2\times2$, it might be easier to use Cayley-Hamilton theorem instead, because fewer terms are involved.

Let $p=\operatorname{tr}(A)+1$ and $q=1-\det(A)$. The difference between the two sides of the inequality then becomes \begin{align} &\det(pA+qI)-\frac34q^2\\ &=p^2\det(A) + pq\operatorname{tr}(A) + q^2 - \frac34q^2\\ &=p^2(1-q) + pq(p-1) + \frac14q^2\\ &=p^2 - pq + \frac14q^2\\ &=\left(p-\frac q2\right)^2. \end{align}