Let $a,b$ be the smaller sides of a right triangle. Let $c$ be the hypothenuse and $h$ be the altitude from the right angle. Fint the maximal value of $\frac{c+h}{a+b}$.
1. **Define the variables and the right triangle:**
   Let $ABC$ be a right triangle with $BC = a$, $CA = b$, and $AB = c$. The right angle is at $C$, so $\angle BCA = 90^\circ$. Let $\angle ABC = \theta$.

2. **Express the sides in terms of $c$ and $\theta$:**
   \[
   a = c \cos \theta, \quad b = c \sin \theta
   \]

3. **Find the altitude $h$ from the right angle to the hypotenuse:**
   \[
   h = a \sin \theta = c \cos \theta \sin \theta
   \]

4. **Formulate the expression for $\frac{c + h}{a + b}$:**
   \[
   \frac{c + h}{a + b} = \frac{c + c \cos \theta \sin \theta}{c \cos \theta + c \sin \theta} = \frac{c(1 + \cos \theta \sin \theta)}{c(\cos \theta + \sin \theta)} = \frac{1 + \cos \theta \sin \theta}{\cos \theta + \sin \theta}
   \]

5. **Introduce a new variable $x = \cos \theta + \sin \theta$:**
   \[
   (\cos \theta + \sin \theta)^2 = 1 + 2 \cos \theta \sin \theta \implies x^2 = 1 + 2 \cos \theta \sin \theta \implies \cos \theta \sin \theta = \frac{x^2 - 1}{2}
   \]

6. **Substitute $\cos \theta \sin \theta$ back into the expression:**
   \[
   \frac{c + h}{a + b} = \frac{1 + \frac{x^2 - 1}{2}}{x} = \frac{1 + \frac{x^2 - 1}{2}}{x} = \frac{2 + x^2 - 1}{2x} = \frac{x^2 + 1}{2x} = \frac{1}{2} \left( x + \frac{1}{x} \right)
   \]

7. **Determine the range of $x$:**
   Since $x = \cos \theta + \sin \theta$ and $\cos \theta, \sin \theta \in [0, 1]$, we know:
   \[
   \sqrt{2} \cos \left( \theta - \frac{\pi}{4} \right) \leq x \leq \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right)
   \]
   Therefore, $1 < x \leq \sqrt{2}$.

8. **Maximize the function $\frac{1}{2} \left( x + \frac{1}{x} \right)$:**
   The function $y = \frac{1}{2} \left( x + \frac{1}{x} \right)$ is increasing in the interval $1 < x \leq \sqrt{2}$. Thus, the maximum value occurs at $x = \sqrt{2}$.

9. **Calculate the maximum value:**
   \[
   \frac{1}{2} \left( \sqrt{2} + \frac{1}{\sqrt{2}} \right) = \frac{1}{2} \left( \sqrt{2} + \frac{\sqrt{2}}{2} \right) = \frac{1}{2} \left( \frac{3\sqrt{2}}{2} \right) = \frac{3\sqrt{2}}{4}
   \]

The final answer is $\boxed{\frac{3\sqrt{2}}{4}}$