Solution:
Consider that f(x)=x2+6x+7=(x+3)2−2. So f(x)≥−2 for real numbers x. Also, f is increasing on the interval [−3,∞).
Therefore
f(f(x))≥f(−2)=−1,f(f(f(x)))≥f(−1)=2,
and
f(f(f(f(x))))≥f(2)=23
Thus, the minimum value of f(f(f(f(x)))) is 23 and equality is obtained when x=−3.