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Find all positive real solutions $(a, b, c, d)$ to the following system of equations:
$$a+b+c+d=12, a b c d=27+a b+a c+a d+b c+b d+c d .$$
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## 1 Answer

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Hint 1: Show that $\sqrt{a b c d} \leq 9$. Hint 2 : Show that $\sqrt{a b c d} \geq 9$

Proof of hint 1: Applying AM-GM, $3=\frac{a+b+c+d}{4} \geq \sqrt{a b c d}$. Since both sides are positive, we may square both sides to get $9 \geq \sqrt{a b c d}$.

$b c+b d+c d \geq 6 \sqrt{a b c d}$. Hence, $(\sqrt{a b c d})^{2}-6 \sqrt{a b c d}-27 \geq 0$, or that $(\sqrt{a b c d}-9)(\sqrt{a b c d}+3) \geq 0 .$ since the second term of the product is always positive, this implies $\sqrt{a b c d}-9 \geq 0 . \square$
gives $a=b=c=d=3$ as the only possible solution. A quick check shows that this is indeed a solution. $\square$
by Diamond (75,914 points)

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