Solutions

It's somewhat difficult to imagine solving the whole
inequality chain directly. Instead, we segment the inequality into
multiple parts which we individually all prove.

That is, we first prove that $\bm{a < \sqrt{ab}}$:

$$\begin{aligned} 0 < a < b & \implies \sqrt{a} < \sqrt{b} && \text{Square root function being strictly increasing} \\ & \implies \sqrt{a} \sqrt{a} < \sqrt{b} \sqrt{a} \\ & \implies a < \sqrt{ab} \end{aligned}$$

Not so fast, as we need to prove that the square roof function is
strictly increasing. Suppose $a, b \in \mathbb{R}: 0 < a < b$, then:

$$\begin{aligned} 0 < a < b & \implies b - a > 0 \\ & \implies \frac{b - a}{\sqrt{b} + \sqrt{a}} > 0 && \text{Difference of squares} \\ & \implies \sqrt{b} - \sqrt{a} > 0 \end{aligned}$$

We then prove that $\bm{\sqrt{ab} < \frac{a + b}{2}}$ by assuming
that the inequality holds and by further developing it to arrive at a
trivially true inequality. This is exactly how one should tackle
exercises when only bewilderment is felt; one should "play around"
with the given inequality/equation until either a clue or an answer
becomes apparent.

$$\begin{aligned} \sqrt{ab} < \frac{a + b}{2} & \implies 2 \sqrt{ab} < a + b \\ & \implies 0 < a + b - 2 \sqrt{ab} && (1) \end{aligned}$$

Most students should be capable of arriving at the inequality above but
it might seem like a dead end. By the inequality, it is obvious that we
must prove that the sum of the terms on the right-hand side is positive.
But how? Fret not, for if you rearrange the aforementioned terms, you
may observe a possible refactoring.

$$\begin{aligned} 0 < a - 2 \sqrt{ab} + b && (1) \end{aligned}$$

Do you see it? It is a complete square! We factor it so as to obtain:

$$\begin{aligned} 0 < { (\sqrt{a} + \sqrt{b}) }^{2} \end{aligned}$$

Now, the inequality above is trivially true since $a, b > 0$.

Finally, we must prove that $\bm{\frac{a + b}{2}} < b$:

$$\begin{aligned} a < b & \implies a + b < 2b \\ & \implies \frac{a + b}{2} < b \end{aligned}$$

Using the property of transitivity, we have thus proven the original
inequality:

$$\begin{aligned} a, b > 0 \implies 0 < a < \sqrt{ab} < \frac{a + b}{2} < b \end{aligned}$$