<s>[INST] Proof $ab-cd \ge 3$ $a$, $b$, $c$, and $d$ are real number
$a\ge b\ge c\ge d$
$a+b+c+d = 13$
$a^2+b^2+c^2+d^2 =43$
Proof that $ab-cd\ge 3$
 [/INST] Hint
$$ (a + b + c + d)^2 = 169 = \sum a^2 + 2 ( ab + bc + cd + da + ac + bd) $$
This gives: $ ab + bc + cd + da + ac + bd = 63 $. We already had $ ab + cd \leq \frac{43}{2} $ from $ (a - b)^2 + (c - d)^2 \geq 0 $.
Use the above results with $$ \left[ (a + b) - (c + d) \right]^2 \geq 0 $$
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