<s>[INST] Prove that if $0Prove that if $0<a<b$, then
$a<\sqrt{ab}<\dfrac{a+b}{2}<b.$
I have seen that the inequality $\sqrt{ab}\leq (a+b)/2$ is verified for all $a,b\geq 0$.
 [/INST] This are $3$ inequalities, and $1$ is already solved. $2$ to go, lets solve them by equivalence transformations:
The first one ($a < \sqrt{ab}$):
\begin{align}
a &< \sqrt{ab} \\
a^2 &< ab \\
a &< b
\end{align}
The last one ($\dfrac{a+b}{2} < b$):
\begin{align}
\dfrac{a+b}{2} &< b \\
a + b &< 2 \cdot b \\
a &< b
\end{align}
</s>