To tackle the problems stated in your notes, let's break them down one by one.

### a) Prove that \( a + b > \sqrt{ab} \) given \( a > 10 \) and \( b > 10 \).

To prove this, we can start by manipulating the inequality:

1. **Square both sides**:
\[
(a + b)^2 > ab
\]

2. **Expand the left side**:
\[
a^2 + 2ab + b^2 > ab
\]

3. **Rearranging gives**:
\[
a^2 + b^2 + ab > 0
\]

Since both \( a \) and \( b \) are greater than 10, their squares and their product will certainly be positive. Hence, \( a^2 + b^2 + ab > 0 \) is true for \( a, b > 10 \).

Thus, \( a + b > \sqrt{ab} \) is established.

### b) Prove that:
\[
\frac{bc}{a} + \frac{ca}{b} + \frac{ab}{c} > a + b + c
\]
given \( a, b, c > 0 \).

This is a classic application of the **Cauchy-Schwarz inequality**, which states that for any positive numbers, the following holds:
\[
(x_1^2 + x_2^2 + x_3^2)(y_1^2 + y_2^2 + y_3^2) \geq (x_1y_1 + x_2y_2 + x_3y_3)^2
\]

Applying Cauchy-Schwarz here:

Set \( x_1 = \frac{b}{a}, x_2 = \frac{c}{b}, x_3 = \frac{a}{c} \) and \( y_1 = a, y_2 = b, y_3 = c \).

Using Cauchy-Schwarz, we get:
\[
\left( \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \right)(a + b + c) \geq (b + c + a)^2
\]
Rearranging gives:
\[
\frac{b}{a} + \frac{c}{b} + \frac{a}{c} > 1
\]

Multiplying by \( a + b + c \) yields:
\[
bc + ca + ab > a^2 + b^2 + c^2
\]

This establishes:
\[
\frac{bc}{a} + \frac{ca}{b} + \frac{ab}{c} > a + b + c
\]

Both inequalities are proven.