If in any triangle ABC, the area $\Delta \le \frac{b^2+c^2}{\lambda}$, then the largest possible numerical value of $\lambda$ is?

Question

$$\frac 12 bc \sin A \le \frac{b^2+c^2}{\lambda}$$ $$bc(\frac 12 \lambda \sin A-2)\le (b-c)^2$$

I really didn’t understand this well, so this pretty much what any rookie would do. What should I do next?

The answer is 4.

Answer 1

We have $$\lambda\sin(\alpha)\le 2\left(\frac{b}{c}+\frac{c}{b}\right)$$ since we have $$\frac{b}{c}+\frac{c}{b}\geq 2$$ we get $$2\left(\frac{b}{c}+\frac{c}{b}\right)\geq 4$$

Answer 2

They are asking what the largest value of $\lambda$ is that makes the inequality valid. It boils down to two issues. First, what is the maximum value of $\sin A$? Secondly, what is the best inequality of the form $bc \leq C(b^2 + c^2)$. The second one is related to the AM GM inequality. Since $\sin A$ is at most $1$ and $C$ is at most ${1 \over 2}$, the best you can do is $\lambda = 4$.