<s>[INST] 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 $60^{\circ}$.
 [/INST] Let $a<\sqrt{b^2-bc+c^2}.$
Thus, $$b^2<ac<c\sqrt{b^2-bc+c^2},$$ which gives
$$(b-c)(b^2(b+c)+c^3)<0$$ or
$$b<c.$$
Also, since $$a(2b-c)>bc,$$ we obtain: $$2b-c>0$$ and
$$\sqrt{b^2-bc+c^2}>a>\frac{bc}{2b-c},$$ which gives $$(b-c)(4b^3-4b^2c+4bc^2-c^3)>0$$ or
$$4b^3-4b^2c+4bc^2-c^3<0.$$
But $$4b^3-4b^2c+4bc^2-c^3=b(2b-c)^2+c^2(2b-c)+c^2b>0,$$
which is a contradiction.
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