This is a classic problem involving geometry and relative speed, often referred to as the "Duck and Fox Problem". Here's the approach:

1. Swim in Circles: The duck begins by swimming in a circle with a radius of $r < R/4$ where $R$ is the radius of the lake. The center of this circle is the same as that of the lake. The fox is four times as fast as the duck. By swimming in a small circle, the duck can start creating an angular separation between itself and the fox.

2. Force the Fox to Run: Since the fox is trying to chase the duck, it will continue to run around, trying to move to the point nearest to the duck but on the shore. As the duck goes around slightly faster, the fox lags behind. Eventually, the duck will be at a point where the fox is directly across the lake from it.

3. Wait for the Right Moment: The duck continues swimming in its smaller circle until the fox is straight across the lake from it. At this point, the duck and fox are separated by half the circumference of the lake, a distance that the fox has to cover to reach the duck if it heads straight for the shore.

4. Head for the Shore: While standing directly opposite, the duck can now swim straight to the shore. Duck has to travel around $R \cdot 3/4$ distance, while the fox has to travel $\pi R$. Given the ratio of their speeds, it will take $3 R$ and $3.14 \cdot R$ units of time respectively. Hence the duck will have a few extra moments to start the flight.

5. Escape: Reaching moments before the fox, the duck can now fly away to safety.

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Limits of $r$

Let us also formalize the limits of the choice of $r$.

It is slightly less than $R/4$, to create a phase lag

$r < R/4$ (to create a phase lag)

Also, in the end, the remaining time to reach the shore should be less for the duck, than the fox.

$\dfrac{(R-r)}{1v} < \dfrac{\pi R}{4v}$ (where $v$ is the speed.)

Solving these two equations gives:

${R} \cdot (1 - \dfrac{\pi}{4}) < r < \dfrac{R}{4}$

Hence, the inner circle can have a radius between $21.5\%$ to $25\%$ of $R$