# Proving a complicated inequality involving integers

Let $a,b,c,d$ be integers such that $$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right) \mod 2$$ $$ad-bc =1$$ $$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \neq \left( \begin{matrix} \pm 1 & 0 \\ 0 & \pm 1 \end{matrix}\right) .$$

Let $(x,y)\in \mathbf{R}^2$ such that $-1\leq x\leq 1$ and $1/2 \leq y\leq 2$.

I highly suspect that $$c^2(y^2+2x^2) + a^2+d^2+2cx(d-a) + \frac{1}{y^2}(b-(d-a+3cx)x)^2 \geq 3.$$

The proof is actually very easy and was obtained after Parsa's answer.

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Where is the connection between your matrix and $(1,2)$? – draks ... Apr 1 '12 at 17:41
You could write more about where this inequality came from. – Beni Bogosel Apr 4 '12 at 10:07

In the case where $a$ and $d$ have the same sign, rewrite your expression as $$c^2(y^2 + x^2) + (cx + (d-a))^2 + 2ad + \frac{1}{y^2}(b-(d-a+3cx)x)^2$$
Since each term is non-negative your inequality follows easily. Perhaps you can adjust this expression so your inequality follows in the case where $a$ and $d$ have opposite signs.
Thank you for this answer! Your observation inspired me to look further and I found a proof of an even stronger inequality. In fact, one simply writes $(cx+(d-a))^2 + 2ad + c^2x^2 = 2c^2x^2+ a^2+d^2+ 2cx(d-a) = 2 ( (cx)^2 + cx(d-a)) + a^2+d^2 = 2( (cx + (d-a)/2)^2 - (d-a)^2/4) + a^2+d^2 = 2(cx+(d-a)/2)^2 - (d-a)^2/2 +a^2+d^2= 2(cx+(d-a)/2)^2 +(a+d)^2/2$. – Harry Apr 9 '12 at 10:07
Now note that $\vert a+d\vert \geq 2$ for all the matrices we are considering. – Harry Apr 9 '12 at 10:09
And of course $c^2y^2 \geq 1$ if $c\neq 0$ (we may assume that $c\neq 0$). All the other terms are nonnegative. – Harry Apr 9 '12 at 10:11