Is product of any isometries in $\mathbb{R}^n$ an isometry?

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I am reading the book Putnam and Beyond by R.Gelca and T.Andreescu, and there is the following problem (#118):

Let $V_1,V_2,...,V_m$ and $W_1,W_2,...,W_m$ be isometries of $\mathbb{R}^n$ ($m,n$ positive integers). Assume that for all $x$ with $||x|| \leq 1$, $||V_i x - W_i x|| \leq 1,\ i = 1,2,...,m$. Prove that: $$ \Big|\Big| \big(\prod_{i=1}^m V_i\big)\ x - \big(\prod_{i=1}^m W_i\big)\ x \Big|\Big| \leq m $$ for all $x$ with $||x|| \leq 1$.

Why the following solution is wrong (I presume that it's wrong because authors used a more sophisticated one and mine is too obvious):

We know, that isometries form a group, so $V = V_2 \cdot ... \cdot V_m$, $W = W_2 \cdot ... \cdot W_m$ are isometries too. That's why for all $x$ such that $||x|| \leq 1$ we have $||Vx|| = ||x|| \leq 1$ and $||Wx|| = ||x|| \leq 1$. But this means that vectors $Vx$ and $Wx$ satisfy the property from the task and we have $||V_1(Vx) - W_1(Wx)|| \leq 1$, which is the desired result (and even stronger).

P.S. If my argument is valid, then what condition should be added to the task so it is not valid anymore?

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You can only conclude $$\lVert V_1(Vx) - W_1(Wx) \rVert = \lVert V_1(Vx) - V_1(Wx) + V_1(Wx) - W_1(Wx) \rVert \le \lVert V_1(Vx) - V_1(Wx) \rVert + \lVert V_1(Wx) - W_1(Wx) \rVert = \lVert Vx - Wx \rVert + \lVert V_1(Wx) - W_1(Wx) \rVert \le\lVert Vx - Wx \rVert + 1.$$

This reduces a product of $m$ isometries to one of $m-1$.

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I agree with your observations but then we should conclude that

$$\|Vx-Wx\|\le \|Vx\|+\|Wx\|\le 2$$