A curious geometric inequality

I am trying to prove the following inequality: $$\tag{*}2 \le \prod_{i=1}^d r_i + \prod_{i=1}^d (|r_i-1|+1)$$ where $r_i \ge 0$.

I came up with a proof using induction (see below) but I wonder if there is a simpler proof.

The goal of the question is to see alternative proofs. Maybe there is a proof with geometric insight?

Motivation:

This inequality comes up when proving a bound on the volume distortion of a linear transformation $A$ in terms of its deviation from being an isometry $\operatorname{dist}(A,\operatorname{SO}_d)$; The idea is that when $A$ is close to $\operatorname{SO}_d$ the distortion should be small. In that context the $r_i$ are the singular values of $A$, and inequality $(*)$ is used to show $$|\det A -1| \le (\operatorname{dist}(A,\operatorname{SO}_d)+1)^d-1$$

My Proof:

We prove by induction over $d$. For $d=1$ the inequality reduces to $$2 \le r+|r-1|+1 \iff 1 \le r+|r-1|.$$ checking the two cases $r \ge 1, r <1$ is trivial.

Assume the inequality holds for $d-1$. Denote $x=\prod_{i=1}^{d-1} r_i,y=\prod_{i=1}^{d-1} (|r_i-1|+1)$. Then, by assumption, we have $$2 \le x+y$$

Note that $y \ge 1$.

Denote $r=r_d$. We need to prove $$2 \le xr+y(|r-1|+1)$$.

The case $r \ge 1$:

$$2 \le xr+y(|r-1|+1) \iff 2 \le r(x+y)$$

This holds since $r \ge 1, x+y \ge 2$.

The case $0 \le r < 1$:

$$2 \le xr+y(|r-1|+1) \iff 2 \le xr+y(2-r)$$

Since $x+y \ge 2 \Rightarrow x \ge 2-y \Rightarrow xr \ge (2-y)r$ (remember $r \ge 0$), so it suffices to prove

$$2 \le (2-y)r+y(2-r)=2(r+y-ry)$$ which holds if and only if

$$1 \le r+y-ry \iff 1-r \le y(1-r)$$ which indeed holds since $y \ge 1,1-r \ge 0$.

Fix an $n\geq0$, and put $\prod_{i=1}^n r_i=:A$, $\>\prod_{i=1}^n \bigl(|r_i-1|+1\bigr)=:B$, and $r_{n+1}=:r$. I claim that $0\leq A\leq B$ and $A+B\geq2$. This is obviously true for $n=0$. Now let $n\rightsquigarrow n+1$. If $r\geq1$ then $A'=Ar$ and $B'=Br$, hence the claim is true for $n+1$ in this case. If $0\leq r\leq1$ then $$A'=Ar,\quad B'=B(2-r)\geq Br\ .$$ This implies $A'\leq B'$ and $$A'+B'=A+B+(1-r)(B-A)\geq2\ .$$ It cannot get simpler than this!
For a "one stroke" proof you can assume $0\leq r_i\leq1$ for all $i$, since the $r_i>1$ do not cause any difficulties. We then write $r_i:=1-x_i$ with $0\leq r_i\leq1$ and have $$A+B=\prod_{i=1}^n(1-x_i)+\prod_{i=1}^n(1+x_i)\geq2\ ,$$ since under distributive multiplication all terms with an odd number of $x_i$'s cancel.