Maximum volume of water one can drink without being burnt A friend asked me the following interesting question:

We have a bottle with 1L water of 20°C and an unlimited supply of boiling water (100°C). A person can only drink water of temperature $\leq 60$°C. What is the maximum amount of water one can drink if we assume waiting doesn't cool the water and we can only drink directly from the bottle?

Note: the temperature of the result of mixing $m_1$ L water of $t_1$°C and $m_2$ L water of $t_2$°C is $\frac{m_1t_1+m_2t_2}{m_1+m_2}$°C
A strategy is to always drink a fixed portion of water from the bottle, add boiling water until the bottle is full again, drink, add, ..., until when adding would make the water exceed 60°C, at which time we add water until we reach 60°C and drink all of it.
Let the portion of the remaining water in the bottle after drinking be $0<r<1$ (i.e. we drink $1-r$ L from the bottle). I could prove that after drinking and adding the $n$-th time, the temperature in the bottle is (°C)
$$t_n=100-80r^n.$$
Therefore we would have drunk $$n=\lfloor-\log_r 2\rfloor$$ times until we reach 60°C, $(1-r)$ L each time. 
Since $\lim_{r \to 1}n(1-r)=\ln 2$, we can drink up to $(1+\ln 2)$ L water together with the remaining whole bottle 60°C water.
My question:
Is this strategy optimal? How can I prove it if so? How can we do better otherwise?
Any help is appreciated, thanks!

If you drink at each step $m_i$ of the bottle and fill up with hot water, you will get,
$$t_n=20\prod_{i=1}^n (1-m_i) + 100\sum_{k\geq 1}^nm_k\prod_{i\geq k+1}(1-m_i) $$
I think, to prove your method, you have to show that this quantity is minimum, for a fixed $n$ and a fixed $0<k<1$ such that $m_1+\cdots+m_n=k$, when $m_1=\dots = m_n$,  which seems to me a rather classical optimisation problem. 
For exemple for $n=2$, 
$$t_1=20+ m_1 80$$
and 
$$t_2= 20+(m_1+m_2-m_1m_2)80 $$
With $m_1+m_2=k$,
  $$t_2= 20+(k-m_1(k-m_1))80 =20+k80-m_1(k-m_1)80,$$
so you need $m_1(k-m_1) $ being maximal (with $0<m_1<k$) so 
$$m_1 =\frac{k}{2} $$
Since $k$ could be any value, it is proven that, for two steps, the minimal temperature is obtain (for a subtracted fixed volume) by two equal portion. 
