Suppose $p, q, r$ are real numbers such that $q = p (4 - p), r = q (4 - q), p = r (4 - r) .$ The maximum possible value of $p + q + r$ is

0

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3

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9

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27

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Solution

The correct option is D $9$
$p + q + r = 4 (p + q + r) - (p^{2} + q^{2} + r^{2}) \Rightarrow p + q + r = \frac{p^{2} + q^{2} + r^{2}}{3}$
Since, $\frac{1}{3} (p + q + r)^{2} \leq (p^{2} + q^{2} + r^{2})$
$∴ \frac{1}{3} (p + q + r)^{2} \leq 3 (p + q + r) \Rightarrow (p + q + r) \leq 9$
If $p = q = r = 3 so, max (p + q + r) = 9$

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