Show that $b^2+c^2-a^2\leq bc$. Let $a,b,c>0$ such that $b<\sqrt{ac}$, $c<\frac{2ab}{a+b}$. Show that $b^2+c^2-a^2\leq bc$.
I tried to construct a triangle with $a,b,c$ and to apply The cosine rule, but I am not sure that it's possible to construct it and also I have no idea how to prove that an angle it's greater than $60^{\circ}$.

Using $ac - b^2 > 0$ and $2ab - (a + b)c > 0$, we have
\begin{align*}
 &bc - (b^2 + c^2 - a^2)\\
 >\,&bc - (b^2 + c^2 - a^2) - (ac - b^2) - [2ab - (a + b)c]\\
 =\,& a^2 - 2ab + 2bc - c^2\\
 =\,&(a - c)(a + c - 2b)\\
 >\,& 0
\end{align*}
where we have used $a > c$ and $a + c > 2b$.
(Explanations:
By AM-GM, we have
$2b < 2\sqrt{ac} \le a+c$.
Also, by AM-GM, we have
$$c < \frac{2ab}{a + b} \le \frac{2ab}{2\sqrt{ab}} = \sqrt{ab} < \sqrt{a \sqrt{ac}}$$
which results in $c < a$.)
We are done.
