Question

If $p=r(4-r)$
$q=p(4-p)$
$r=q(4-q)$
where $p\ne q\ne r\ne 0$
Then find minimum value of $(p+q+r)$

Answer 1

$\begin{array}{c}p=r(4-r)\\ q=p(4-p)\\ r=q(4-q)\\ p+q+r=r(4-r)+p(4-p)+q(4-q)\\ \Rightarrow p+q+r=4r-r^2+4p-p^2+4q-q^2\\ \Rightarrow p+q+r=4(p+q+r)-(p^2+q^2+r^2)\\ \Rightarrow 3(p+q+r)=(p^2+q^2+r^2)\\ \Rightarrow 3(p+q+r)=(p+q+r{)}^2-2(pq+qr+rp)\\ Let,p+q+r=x\\ \Rightarrow 3x=x^2-2(pq+qr+rp)\\ \Rightarrow x^2-3x-2(pq+qr+rp)=0\end{array}$
which is quadratic
Minimum value is given by $\frac{-b}{2a}=\frac{-(-3)}{2\left(1\right)}=\frac{3}{2}$