Inequality involving sums and products (related to multi-stage rockets) Let $a_i$ and $b_i$ be strictly positive real numbers for $i=1,\ldots, n$. I wonder whether the following inequality holds in general or does not:$$\frac{\sum_{i=1}^n a_i+\sum_{i=1}^n b_i}{\sum_{i=1}^n a_i}<\prod_{k=1}^n\frac{\sum_{i=1}^k a_i+\sum_{i=1}^k b_i}{\sum_{i=1}^k a_i+\sum_{i=1}^{k-1} b_i}.$$
I have not been able to prove it in general, but I am not finding a countexample [EDIT: the non-strict inequality holds for $n=1$, as pointed out by Oskar, whom I thank], either, therefore I am inclined to suspect that it is true. How can it be proved, if it is? I thank you so much for any answer!

Background: the question has arisen in my mind because I read in a physics text that the efficience of a one stage rocket, whose velocity variation can be expressed, in absence of external forces and for constant speed $\mathbf{v}_{gr}$ and rate of expulsion $\frac{dm}{dt}$ of the gas, by the Ciolkovskij rocket equation
$$\mathbf{v}_f-\mathbf{v}_i=\mathbf{v}_{gr}\ln\Big(\frac{m_f}{m_i}\Big)$$
where $m_i$ and $m_f$ respectively are the initial (rocket and gas) and final (rocket only) masses, can be increased by using multi-stage rockets.
I would say that, if we call $a_i$ the mass of stage $i$ and $b_i$ the mass of the gas contained in it, the above speed variation, considering a one stage rocket made of $n$ "stage-like" parts, which expels the gas at once, can be written as $$\Delta v_{\text{one stage rocket}}=-\|\mathbf{v}_{gr}\|\ln\frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n a_i+\sum_{i=1}^n b_i}=\|\mathbf{v}_{gr}\|\ln\frac{\sum_{i=1}^n a_i+\sum_{i=1}^n b_i}{\sum_{i=1}^n a_i}$$ while, if we discharge the gas contained in each stage $i$ at speed $\mathbf{v}_{gr}$ and then stage $i$ is detached, I would say that the variation is (letting $\sum_{i=1}^0 b_i:=0$) $$\Delta v_{\text{multi-stage rocket}}=\sum_{k=1}^n \|\mathbf{v}_{gr}\| \ln\frac{\sum_{i=1}^k a_i+\sum_{i=1}^k b_i}{\sum_{i=1}^k a_i+\sum_{i=1}^{k-1} b_i}$$and $\Delta v_{\text{one stage rocket}}<\Delta v_{\text{multi-stage rocket}}$ is equivalent to the inequality that is the object of the question.
 A: Let denote:
$$ s_k=\sum_{i=1}^k a_i\ \ \ \ \ t_k=\sum_{i=1}^kb_i$$
hence the inequality to prove is :

$$P(n)\ \ \ \ \ \ \ \ \ \ 1+\frac{t_n}{s_n}\leq \prod_{k=1}^{n} \frac{s_k+t_k}{s_k+t_{k-1}}=\prod_{k=1}^{n} \left(1+\frac{b_k}{s_k+t_{k-1}}\right)$$

Proof (by induction on $n$)
Basis step $P(1)$ is obviously true because:
$$(1+\frac{b}{a})\leq \prod_{i=1}^1 \left(1+\frac{b_i}{s_i+t_{i-1}}\right)=(1+\frac{b}{a})$$
Induction step Let's assume that $P(n)$ is true, given $a_1,\cdots,a_{n+1},b_1,\cdots,b_{n+1}$ strictly positive numbers we have:$$s_nt_n<s_{n+1}t_{n}$$ 
because $0<a_{n+1}$and this is equivalent to (by adding $s_ns_{n+1}$ in both sides of the inequality and multiplying by $(t_{n+1}+s_{n+1})$) $$(s_{n+1}+t_{n+1})(s_{n+1}+t_n)s_n<s_{n+1}(s_n+t_n)(s_{n+1}+t_{n+1}) $$ 
which  implies that:
$$1+\frac{t_{n+1}}{s_{n+1}}<\left(1+\frac{t_n}{s_n}\right)\left(1+\frac{b_{n+1}}{s_{n+1}+t_n}\right)  $$
hence using induction hypothesis:
$$ 1+\frac{t_{n+1}}{s_{n+1}}< \left(1+\frac{t_n}{s_n}\right)\cdot \left(1+\frac{b_{n+1}}{s_{n+1}+t_n}\right) \leq \prod_{k=1}^{n+1} \left(1+\frac{b_k}{s_k+t_{k-1}}\right)$$
and the induction terminates.
The inequality is strict for $n\geq 2$ because in the induction step even if $P(n)$ is not strict, it implies that $P(n+1)$ strict, so the only non strict inequality is $P(n)$
