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 $60^{\circ}$.

Another solution.
First, we find $c<\frac{2ab}{a+b}\leq \frac{2\left(\frac{a+b}{2}\right)^2}{a+b}=\frac{a+b}{2}$ by applying AM-GM to $ab$. We also find $b<\sqrt{ac}\leq \frac{a+c}{2}$.
Now if $(b+c)\leq a$, this implies $(b+c)^2\leq a^2 \Rightarrow b^2+c^2-a^2\leq -2bc<bc$, and we are done.
If not, then $b+c>a$ we'll have a triangle $\triangle ABC$. By the law of cosine,
$$a^2 = b^2+c^2-2bc\cos{A} \Rightarrow b^2+c^2-a^2=2bc\cos{A}$$ we only need to consider cases where $\angle A<60^{\circ}$ which would make our statement false.
Now $\angle A<60^{\circ}$ implies $a$ can't be the longest side, and the condition $b<\sqrt{ac}$ implies $b$ can't be the longest side. Therefore $c$ must be the longest side. However $c<\frac{a+b}{2}$, so that contradicts the fact we have a triangle.
