If $\Delta$ is the area of the triangle $ABC$, Show that $\Delta \le \tfrac{b^2+c^2}{4}$.

Question

I thought of using ravi's substitution and using of herons formula to solve this but unable to reach the final answer.

Answer 1

I think, it's better the following:

By AM-GM $$\frac{b^2+c^2}{4}\geq\frac{bc}{2}\geq\frac{bc\sin\alpha}{2}=\Delta.$$