If $0<a<b$, prove that $a<\sqrt{ab}<\frac{a+b}{2}<b$ [duplicate]

If $0<a<b$, prove that $a<\sqrt{ab}<\frac{a+b}{2}<b$

So far I've got:

$a<b$

$a^2<ba$

$a<\sqrt{ab}$

And:

$a<b$

$a+b<2b$

$\frac{a+b}{2}<b$

So I need to prove that $\sqrt{ab}<\frac{a+b}{2}$

How can I do it?

I am going to show you, via screenshots (too much effort to try to typeset everything again on here), a chain of inequalities I proved. Your problem is basically the same but even a littler easier.

For my problem, I was given that $a,b\in\mathbb{R^+}$ and $a\leq b$, and I was tasked with proving that $$ a\leq \frac{2ab}{a+b} \leq\sqrt{ab}\leq\frac{a+b}{2}\leq\sqrt{\frac{a^2+b^2}{2}}\leq b. $$ Your problem may be easily solved by adapting my work that appears below.

If you adapt things correctly, your inequalities will easily fall out.

That's a private case of AM-GM: $$(\sqrt a-\sqrt b)^2>0\Leftrightarrow a+b-2\sqrt {ab}>0\Leftrightarrow\frac{a+b}{2}>\sqrt{ab}$$