Answer
See the detailed answer below.
Work Step by Step
We know that the average speed is given by
$$v_{\rm avg}=\dfrac{\Delta x}{\Delta t}$$
And for a particle that undergoes a simple harmonic motion, during a one-half period $T/2$, it moves from $+A\rightarrow -A$.
This means that $\Delta x=-A-(+A)=-2A$ and we can neglect the negative sign here.
Thus,
$$v_{\rm avg}=\dfrac{2A}{\frac{1}{2}T}=\dfrac{4A}{T}$$
We know that the period $T$ is given by $\omega=2\pi f=2\pi/T$
Hence, $T=2\pi/\omega$
$$v_{\rm avg} =\dfrac{4A}{\frac{2\pi}{\omega}}=\dfrac{2 A\omega}{\pi }$$
Recalling that $v_{\rm max}=A\omega$,
$$v_{\rm avg} =\dfrac{2 v_{\rm max}}{\pi }$$
Therefore,
$$\boxed{v_{\rm max}=\dfrac{\pi v_{\rm avg}}{2}}$$