# 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 $ab-cd\ge 3$

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i don't get ab-cd>=3 ; but i get ab-cd>= -43/2 with the quadratic number theorem – graham sturdy Mar 20 '13 at 5:39
What is the "quadratic number theorem", and how do you use it to get your result? Where does this problem come from? Why is it of interest? – Gerry Myerson Mar 20 '13 at 5:40
i mean (a-b)^2 + (c-d)^2 >= 0 because quadratic of real numbers more than or equal to zero – graham sturdy Mar 20 '13 at 5:43

$$(a + b + c + d)^2 = 169 = \sum a^2 + 2 ( ab + bc + cd + da + ac + bd)$$ 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 $$\left[ (a + b) - (c + d) \right]^2 \geq 0$$