In isosceles right-angled triangle $ABC$, $CA = CB = 1$. $P$ is an arbitrary point on the sides of $ABC$. Find the maximum of $PA \cdot PB \cdot PC$.
Given an isosceles right-angled triangle \(ABC\) with \(CA = CB = 1\), we need to find the maximum value of \(PA \cdot PB \cdot PC\) where \(P\) is an arbitrary point on the sides of \(\triangle ABC\).

1. **Case 1: \(P \in BC\)**

   Let \(PC = x\). Then \(PB = 1 - x\).

   Using the Pythagorean theorem in \(\triangle CPA\):
   \[
   PA^2 = PC^2 + CA^2 = x^2 + 1 \implies PA = \sqrt{x^2 + 1}
   \]

   Therefore, the product is:
   \[
   PA \cdot PB \cdot PC = x(1 - x)\sqrt{x^2 + 1}
   \]

   We need to maximize the function \(h(x) = x(1 - x)\sqrt{x^2 + 1}\) for \(x \in [0, 1]\).

2. **Case 2: \(P \in AB\)**

   Let \(PH \perp AC\) with \(H \in AC\). Let \(CH = x\). Then \(HA = PH = 1 - x\).

   Using the Pythagorean theorem in \(\triangle CPH\):
   \[
   PC^2 = CH^2 + PH^2 = x^2 + (1 - x)^2 = x^2 + 1 - 2x + x^2 = 2x^2 - 2x + 1
   \]

   Since \(\triangle PHA\) is a 45-45-90 triangle:
   \[
   PA = \sqrt{2} \cdot HA = \sqrt{2}(1 - x)
   \]
   Similarly,
   \[
   PB = \sqrt{2} \cdot x
   \]

   Therefore, the product is:
   \[
   PA \cdot PB \cdot PC = 2x(1 - x) \sqrt{2x^2 - 2x + 1}
   \]

   To find the maximum, we define:
   \[
   f(x) = 2x(1 - x), \quad g(x) = 2x^2 - 2x + 1
   \]

   The product function is:
   \[
   F(x) = f(x) \sqrt{g(x)}
   \]

   Notice that:
   \[
   g(x) + f(x) = 2x^2 - 2x + 1 + 2x(1 - x) = 1 \implies g(x) = 1 - f(x)
   \]

   Differentiating \(F(x)\):
   \[
   F'(x) = f'(x) \sqrt{g(x)} + f(x) \frac{g'(x)}{2\sqrt{g(x)}}
   \]
   \[
   F'(x) = f'(x) \sqrt{g(x)} - f(x) \frac{f'(x)}{2\sqrt{g(x)}}
   \]
   \[
   F'(x) = f'(x) \left[\sqrt{g(x)} - \frac{f(x)}{2\sqrt{g(x)}}\right]
   \]
   \[
   F'(x) = f'(x) \left[\frac{2g(x) - f(x)}{2\sqrt{g(x)}}\right]
   \]
   \[
   F'(x) = f'(x) \left[\frac{2(1 - f(x)) - f(x)}{2\sqrt{g(x)}}\right]
   \]
   \[
   F'(x) = f'(x) \left[\frac{2 - 3f(x)}{2\sqrt{g(x)}}\right]
   \]

   Since \(\frac{2 - 3f(x)}{2\sqrt{g(x)}}\) is always positive, we have:
   \[
   F'(x) = 0 \implies f'(x) = 0
   \]

   Solving \(f'(x) = 2 - 4x = 0\):
   \[
   x = \frac{1}{2}
   \]

   At \(x = \frac{1}{2}\):
   \[
   f\left(\frac{1}{2}\right) = \frac{1}{2}, \quad g\left(\frac{1}{2}\right) = \frac{1}{2}
   \]

   Therefore:
   \[
   F\left(\frac{1}{2}\right) = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}
   \]

3. **Comparison of \(h(x)\) and \(F(x)\)**

   We need to show \(h(x) < F(x)\) for \(x \in (0, 1)\):
   \[
   h(x) < F(x) \implies x(1 - x)\sqrt{x^2 + 1} < 2x(1 - x) \sqrt{2x^2 - 2x + 1}
   \]
   Dividing by the positive \(x(1 - x)\):
   \[
   \sqrt{x^2 + 1} < 2 \sqrt{2x^2 - 2x + 1}
   \]
   Squaring both sides:
   \[
   x^2 + 1 < 4(2x^2 - 2x + 1)
   \]
   Simplifying:
   \[
   x^2 + 1 < 8x^2 - 8x + 4
   \]
   \[
   0 < 7x^2 - 8x + 3
   \]

   The discriminant of \(7x^2 - 8x + 3\) is:
   \[
   D = 8^2 - 4 \cdot 7 \cdot 3 = 64 - 84 = -20 < 0
   \]

   Therefore, \(7x^2 - 8x + 3\) is always positive, confirming \(h(x) < F(x)\) for \(x \in (0, 1)\).

Hence, the maximum value of \(PA \cdot PB \cdot PC\) for all possible positions of \(P\) is \(F\left(\frac{1}{2}\right) = \frac{\sqrt{2}}{4}\).

The final answer is \(\boxed{\frac{\sqrt{2}}{4}}\).