Let $c$ be the length of the hypotenuse of a right angle triangle whose two other sides have lengths $a$ and $b$. Prove that $a+b\le c\sqrt{2}$. When does the equality hold?
1. By the Pythagorean Theorem, we know that for a right-angled triangle with sides \(a\), \(b\), and hypotenuse \(c\), the following equation holds:
   \[
   c^2 = a^2 + b^2
   \]

2. We will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states that for any non-negative real numbers \(x\) and \(y\):
   \[
   \frac{x + y}{2} \geq \sqrt{xy}
   \]
   Applying AM-GM to \(a^2\) and \(b^2\), we get:
   \[
   \frac{a^2 + b^2}{2} \geq \sqrt{a^2 b^2}
   \]
   Simplifying, we have:
   \[
   \frac{a^2 + b^2}{2} \geq ab
   \]
   Multiplying both sides by 2, we obtain:
   \[
   a^2 + b^2 \geq 2ab
   \]

3. Since \(c^2 = a^2 + b^2\), we can substitute \(c^2\) into the inequality:
   \[
   c^2 \geq 2ab
   \]

4. Adding \(c^2\) to both sides of the inequality \(c^2 \geq 2ab\), we get:
   \[
   c^2 + c^2 \geq 2ab + c^2
   \]
   Simplifying, we have:
   \[
   2c^2 \geq a^2 + 2ab + b^2
   \]

5. Notice that \(a^2 + 2ab + b^2\) is a perfect square:
   \[
   a^2 + 2ab + b^2 = (a + b)^2
   \]
   Therefore, the inequality becomes:
   \[
   2c^2 \geq (a + b)^2
   \]

6. Taking the square root of both sides, we obtain:
   \[
   \sqrt{2c^2} \geq a + b
   \]
   Simplifying, we get:
   \[
   c\sqrt{2} \geq a + b
   \]
   or equivalently:
   \[
   a + b \leq c\sqrt{2}
   \]

7. Equality holds when \(a = b\). In this case, the triangle is a right isosceles triangle, and the hypotenuse \(c\) is given by:
   \[
   c = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}
   \]
   Thus, \(a + b = a + a = 2a\) and \(c\sqrt{2} = a\sqrt{2} \cdot \sqrt{2} = 2a\), confirming that equality holds when \(a = b\).

\(\blacksquare\)

The final answer is \( \boxed{ a + b \leq c\sqrt{2} } \)