$1+(1+2b_1)(1+2b_2)\cdots(1+2b_n)\geq 2(1+b_1)(1+b_2)\cdots(1+b_n)$ Prove the statement by induction.
$\forall\; b_1,b_2,...,b_n \in [0,+\infty\rangle\;\;\&\;\; \forall n\in N$:
$$1+\underbrace{(1+2b_1)(1+2b_2)\cdot...\cdot(1+2b_n)}_f\geq 2\underbrace{(1+b_1)(1+b_2)\cdot...\cdot(1+b_n)}_g$$
The base: $n=1$
$$1+(1+2b_1)\geq 2(1+b_1)$$
For $n=1$ we have equality.
Assumption: the statement is true for an arbitrary $n\in \mathbb N.$
I wanted to use some inequality and apply it for the whole expression in the step. I am not sure whether it has any sense to divide the right side of the inequality although all the $(1+b_i)$ factors are greater or equal to 1. Just in case I wrote:
$$\frac{1+2b_i}{1+b_1}=2-\frac{1}{1+b_i},\;\;\; \inf{\Bigg(2-\frac{1}{1+b_i}}\Bigg)=1,\; b_i=0$$
Step: $f\geq g\implies1+f\geq g\implies1+f\cdot b_{n+1}\geq g\cdot b_{n+1}$
Is this correct?
 A: I don’t think you’re proof is correct. Additionally, there is some room for improvement with the language of the proof too.
Using $\inf$ is slightly confusing, since $\inf$ is normally only used when dealing with an infinite number of elements. If you’re dealing with a finite number of elements (and there are a finite number of $b_i$), $\min$ is far more appropriate.
This said, I can see why you used $\inf$ though: you just want to say that provided $b_i \geq 0$, the lowest $2 - \frac{1}{1 + b_i}$ can be is $1$ (And so in this case you’re consider the $\inf$ of $2 - (1 + x)^{-1})$ over non-negative $x$). In such as case I’d recommend just saying “Since $b_i \geq 0$, $2 - \frac{1}{1 + b_i} \geq 2 - 1 = 1$.”
Assuming by $f$ and $g$ you mean $(1 + 2b_1)\cdots (1 + 2b_n)$ and $2(1 + b_1) \cdots (1 + b_n)$ (this should be stated!), the proof breaks down because you want to show that
$1 + f(1 + 2b_{n + 1}) \geq g (1 + b_{n + 1})$
is true if
$ 1 + f \geq g$
which you know is true by the induction assumption. However, when you divide both sides by $1 + b_{n+1}$ you also need to divide the $1$ by $1 + b_{n + 1}$, and you don’t know that $1/(1 + b_{n + 1}) \geq 1$ (in fact, this is only true when $b_{n+1} = 0$!)
As such I’d recommend just trying to solve the problem by expanding out $1 + f(1 + 2b_{n + 1})$ (i.e write $1 + f(1 + 2b_{n + 1}) = 1 + f + 2fb_{n + 1}$ and show that $1 + f + 2fb_{n+1} \geq g(1 + b_{n + 1})$. I was able to do this in a couple of lines from that point.
