Question

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?

• A. $0$
• B. $3$
• C. $9$
• D. $27$

Answer 1

The correct option is C $9$

$q=p(4-p)$ ...(i)

$r=q(4-q)$ ...(ii)

$p=r(4-r)$ ...(iii)

multiplying by $pr$ in (i) by $pq$ in (ii) and by $qr$ in (iii).we get

$pqr=p^{2}r(4-p)$ ...(iv)

$pqr=q^{2}p(4-q)$ ...(v)

$pqr=r^{2}q(4-r)$ ...(vi)

multiplying eq. (iv),(v)and(vi)

$(pqr{)}^3=(pqr{)}^3(4-p)(4-q)(4-r)$

$1=(4-p)(4-q)(4-r)$ ...(vii)

we know that Arithmatic Mean is always greater or equal to Geometric Mean.

$\frac{(4-p)+(4-q)+(4-r)}{3}\ge \sqrt[3]{3(4-p)(4-q)(4-r)}$

from eq(vii)

$\frac{(4-p)+(4-q)+(4-r)}{3}\ge \sqrt[3]{31}$

$\frac{(4-p)+(4-q)+(4-r)}{3}\ge 1$

$\frac{(12-(p+q+r)}{3}\ge 1$

$12\ge (p+q+r)+3$

$12-3\ge (p+q+r)$

$9\ge (p+q+r)$

i.e. $p+q+r\le 9$

so maximum value of $p+q+r$ is $9$.