IRL_ABooklet_2023

Solution 1. Note that $$\begin{array}{rl}(ad+bc+ca-db{)}^2& =(a(c+d)+b(c-d){)}^2\\ & =a^2(c+d{)}^2+2ab(c^2-d^2)+b^2(c-d{)}^2\\ & \le (c+d{)}^2+2ab(c^2-d^2)+(c-d{)}^2\\ & =2(c^2+d^2)+2ab(c^2-d^2)\\ & =2[(1+ab)c^2+(1-ab)d^2]\\ & \le 2[(1+ab)+(1-ab)](\text{since }-1\le ab\le 1)\\ & =4,\end{array}$$ whence $$|ad+bc+ca-db|\le 2.$$

When $a^2\ne 1$, equality can only occur when we are in case (iii), hence $b^2=1$. Then $ab\ne \pm 1$ and equality in the second inequality occurs iff $c^2=d^2=1$. When $b^2\ne 1$, equality can only occur when we are in case (ii), hence $a^2=1$. Then $ab\ne \pm 1$ and equality in the second inequality occurs iff $c^2=d^2=1$. When $a^2=b^2=1$, we either have $a=b=\pm 1$ and then $ab=1$, in which case equality in the second inequality occurs iff $c^2=1$, or we have $a=-b=\pm 1$, hence $ab=-1$, and equality in the second inequality occurs iff $d^2=1$.

Equality holds in the first inequality iff we are in one of the following four cases: (i) $a^2=b^2=1$, or (ii) $a^2=1,c=d$, or (iii) $b^2=1,c=-d$, or (iv) $c=d=0$. When $c=d=0$, the second inequality is strict, so we are not interested in case (iv). --- In summary, equality occurs iff one of the following holds: $$\begin{array}{rl}a& =b\text{and}{a}^2=b^2=c^2=1,\text{ or}\\ c& =-d\text{and}{b}^2=c^2=d^2=1.\end{array}$$

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Alternative solution.

Solution 2. Note that if $x$, $y$ are real numbers then $$|x+y|+|x-y|=2\max (|x|,|y|).$$ For $$\begin{array}{rl}(|x+y|+|x-y|{)}^2& =(x+y{)}^2+2|x+y||x-y|+(x-y{)}^2\\ & =2(x^2+y^2)+2|x^2-y^2|\\ & =4\max (x^2,y^2)\\ & =4(\max (|x|,|y|){)}^2.\end{array}$$ Hence $$\begin{array}{rl}|ad+bc+ca-db|& =|a(c+d)+b(c-d)|\\ & \le |a||c+d|+|b||c-d|\\ & \le |c+d|+|c-d|\\ & =2\max (|c|,|d|)\\ & \le 2.\end{array}$$

The case of equality can be determined in a way similar to Solution 1.