This problem is essentially an estimation problem involving Chebyshev and least-squares approximations.
Suppose we have two $m$-dimensional vectors, $a$ and $b$, where it is known that $\lVert a \rVert_{\infty}$ (Chebyshev) is a better estimate to some estimation problem than $\lVert b \rVert_{\infty}$ (least-squares). We want to show that
\begin{equation} \lVert b \rVert_{\infty} \leq \sqrt{m}\lVert a \rVert_{\infty} \end{equation}
using the fact that for all $z$,
\begin{equation} \frac{1}{\sqrt{m}} \lVert z \rVert_2 \leq \lVert z \rVert_{\infty} \leq \lVert z \rVert_2 \quad (*) \end{equation}
In this case, I happen to have the solution which goes as:
\begin{equation} \lVert a \rVert_{\infty} \geq \frac{1}{\sqrt{m}} \lVert a \rVert_2 \geq \frac{1}{\sqrt{m}} \lVert b \rVert_2 \geq \frac{1}{\sqrt{m}} \lVert b \rVert_{\infty} \end{equation} and the result follows. Here, the first and the last inequality follow from $(*)$. What is strange to me is the middle inequality.
I begin by describing my work so far involving the middle inequality. From the problem formulation, it is implied that
\begin{equation} \lVert a \rVert_{\infty} = \gamma\lVert b \rVert_{\infty} \end{equation}
for some $\gamma\geq 1$. Then, using $(*)$, we can provide lower and upper bounds to this identity as
\begin{equation} \frac{1}{\sqrt{m}} \lVert a \rVert_2 \leq \lVert a \rVert_{\infty} = \gamma\lVert b \rVert_{\infty} \leq \gamma\lVert b \rVert_2 \end{equation}
Since $m\geq 1$, we see that we can tighten the lower bound above (in $m$) by choosing $m=1$. Similarly, we can tighten the upper bound below (in $\gamma$) by choosing $\gamma=1$. It follows that:
\begin{equation} \lVert a \rVert_2 \leq \lVert b \rVert_2 \end{equation} which is contrary to the middle inequality in the provided solution.
Where does my thinking fail? I am not convinced the solution is correct, but I'd like to be sure. Anyway, your help is kindly appreciated.