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 $60^{\circ}$.

Both conditions and the hypothesis are homogenous, so we can scale the variables until $a=1$. Then the problem is equivalent to the implication:
$$b<\sqrt{c}\land c<\frac{2b}{1+b}\Rightarrow b^2+c^2\le1+bc.$$
New conditions are more convenient and can be rewritten to:
$$b^2<c<\frac{2b}{1+b}.$$
Let's devide the condition by $b$ and the hypothesis by $b^2$, the problem is equivalen to:
$$b<\frac{c}{b}<\frac{2}{1+b}\Rightarrow1+\Big(\frac{c}{b}\Big)^2\le\frac{c}{b}+\frac{1}{b^2}.$$
Let's introduce $r:=\frac{c}{b}$. The problem is equivalent to:
$$b<r<\frac{2}{1+b}\Rightarrow1-r+r^2\le\frac{1}{b^2}.$$
The condition implies:
$$b<\frac{2}{1+b}\Rightarrow b+b^2<2\Rightarrow\Big(b+\frac{1}{2}\Big)^2<\frac{9}{4}\Rightarrow b+\frac{1}{2}<\frac{3}{2}\Rightarrow b<1.$$
Returning the limitation $b>0$ together with the $\frac{2}{1+b}$ being decreasing with the supremum at $b=0$ we get:
$$0<b<1\land0<r<2.$$
The case $r\le1$ is easy:
$$1-r+r^2=\frac{3}{4}+\Big(\frac{1}{2}-r\Big)^2\le\frac{3}{4}+\max_{0<r\le1}\Big|\frac{1}{2}-r\Big|^2=\frac{3}{4}+\Big(\frac{1}{2}\Big)^2=1=\frac{1}{1}<\frac{1}{b^2}.$$
For $r>1$ we have
$$\frac{d}{dr}(1-r+r^2)=2r-1>1>0,$$
so the expression $1-r+r^2$ increases with $r$ and we have
$$1-r+r^2<1-\frac{2}{1+b}+\Big(\frac{2}{1+b}\Big)^2=\frac{3+b^2}{(1+b)^2}=\frac{1}{b^2}\frac{(3+b^2)b^2}{(1+b)^2}=$$
$$\frac{1}{b^2}\frac{b^2+2b^2+b^4}{(1+b)^2}<\frac{1}{b^2}\frac{1+2b+b^2}{(1+b)^2}=\frac{1}{b^2}.$$
