Let $ABC$ be a right triangle with right angle at $C$. Let the legs be $AC = x$ and $BC = y$, and hypotenuse $AB = h$. Let $P$ be a point on the hypotenuse $AB$ at distances $a$ and $b$ from sides $AC$ and $BC$ respectively.

Step 1: Setting up the relation

Using coordinate geometry, place the triangle with $C$ at origin, $A$ on x-axis at $(x, 0)$, and $B$ on y-axis at $(0, y)$. A point $P$ on hypotenuse $AB$ can be written as $P = (x(1-t), yt)$ where $t \in [0, 1]$.

Distance from $P$ to x-axis $= yt = a$

Distance from $P$ to y-axis $= x(1-t) = b$

Therefore, $t = a/y$ and $1-t = b/x$.

Since $t + (1-t) = 1$, we get:

$$\frac{a}{y} + \frac{b}{x} = 1 \quad \text{... (1)}$$

Also, by Pythagoras theorem:

$$h^2 = x^2 + y^2 \quad \text{... (2)}$$

Step 2: Expressing $h$ in terms of one variable

From equation (1): $ax + by = xy$

Solving for $y$:

$$y = \frac{ax}{x - b}$$

Substituting in equation (2):

$$h^2 = x^2 + \frac{a^2 x^2}{(x - b)^2}$$

Step 3: Finding minimum using derivative

Let $f(x) = h^2 = x^2 + \dfrac{a^2 x^2}{(x - b)^2}$

$$f'(x) = 2x + \frac{2a^2 x}{(x - b)^2} - \frac{2a^2 x^2}{(x - b)^3}$$

For minimum, $f'(x) = 0$:

$$x(x - b)^3 + a^2 x(x - b) = a^2 x^2$$

Simplifying:

$$(x - b)^3 = a^2 b$$

Therefore:

$$x - b = (a^2 b)^{1/3} = a^{2/3} \cdot b^{1/3}$$

So:

$$x = b + a^{2/3} \cdot b^{1/3} = b^{1/3}(a^{2/3} + b^{2/3})$$

By symmetry:

$$y = a^{1/3}(a^{2/3} + b^{2/3})$$

Step 4: Calculating minimum hypotenuse

$$h^2 = x^2 + y^2$$ $$= b^{2/3}(a^{2/3} + b^{2/3})^2 + a^{2/3}(a^{2/3} + b^{2/3})^2$$ $$= (a^{2/3} + b^{2/3})^3$$

Therefore,

$$\therefore \quad \boldsymbol{h = (a^{2/3} + b^{2/3})^{3/2}}$$

This is the minimum length of the hypotenuse.