Problem 946

Consider that $f(x)=x^{2}+6 x+7=(x+3)^{2}-2$. So $f(x) \geq-2$ for real numbers $x$. Also, $f$ is increasing on the interval $[-3, \infty)$. Therefore $$f(f(x)) \geq f(-2)=-1$$ $$f(f(f(x))) \geq f(-1)=2$$ and $$f(f(f(f(x)))) \geq f(2)=23$$ Thus, the minimum value of $f(f(f(f(x))))$ is 23 and equality is obtained when $x=-3$.