<s>[INST] Minimise a sum of unknown function. 
Let $f(x)$ be a real valued function such that $f(x) > 0\ \  \forall x
 \in \Bbb R$. $f(x)$ is symmetrical about $x = 2$ and $x = 4$.  If
  $\displaystyle S = \sum^{50}_{r= 1} f(r + 4)$ and
  $\displaystyle\prod^{50}_{r = 1}f(r) = 2^{50}$. Find the minimum value
  of $S$.

$$\begin{align}S &= \sum^{46}_{r = 1} f(r + 4) + f(51) + f(52) + f(53) + f(54)\\  &= \sum^{50}_{r=1} f(r) + f(51) + f(52) + f(53) + f(54) - \left(\  f(1) + f(2) + f(3) + f(4)\ \right)\\&\ge 50 \times\left(\prod^{50}_{r=1} f(r)\right)^{1/50} + f(51) + f(52) + f(53) + f(54) - \left(\  f(1) + f(2) + f(3) + f(4)\ \right)\\&= 100 + f(51) + f(52) + f(53) + f(54) - \left(\  f(1) + f(2) + f(3) + f(4)\ \right) \end{align}$$
Now I don't have any idea how to eliminate the remaining unknowns from the above equation. I suspect that $f(51) + f(52) + f(53) + f(54) =\  f(1) + f(2) + f(3) + f(4)$ and $100$ is the answer. I guess we have to use $f(4+ x) = f(4 -x)$. Any idea how to eliminate these unknowns ?
 [/INST] From the symmetry at $x=4$, $f(5) = f(3)$ and from the symmetry at $x=2$, $f(3)=f(1)$. Therefore,  $f(1)=f(3)=f(5)$. Similarly, we can show that for $\forall r:$
$$f(r+2)=f(r)$$
Consequently,
$$S=25 (f(1)+f(2))$$
Also, note that $\prod_{r=1}^{50}f(r)=(f(1)f(2))^{25}=2^{50}$. Thus,
$$f(1)f(2)=4$$
Therefore, $S$ is minimum when $f(1)=f(2)=2$, and thus $\min{(S)}=100$.
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