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$a$, $b$, $c$, 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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What have you tried? –  ᴊ ᴀ s ᴏ ɴ Mar 20 '13 at 5:37
    
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

1 Answer 1

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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