Let a,b and c be the side lengths of triangle ABC respectively. If the perimeter of $\Delta$ABC is 7, and that $\cos A=-\frac{1}{8}$, find the greatest value of $b*c$.
This is how I start the solution:
$$a+b+c=7 \implies b+c=7-a,\quad \cos A=-\frac{1}{8}\\ a^2=b^2+c^2-2bc\cdot \cos A\\ \implies a^2=b^2+c^2+\frac{bc}{4} \\ \implies a^2=(7-a)^2-2bc+\frac{bc}{4} \\ \implies bc=4(7-2a)$$
On
Apply a.m.-g.m.-inequality on $\sqrt{bc}$.
$$\frac{(7-a)^2}{4}=\left(\frac{b+c}{2}\right)^2 \ge bc = 4(7-2a)$$ $$49-14a+a^2 \ge 16(7 - 2a)$$ $$a^2+18a-63 = (a+21)(a-3)\ge 0$$ $$a \le -21 \text{(rejected) or } a \ge 3$$ $a\mapsto \dfrac{(7-a)^2}{4}$ is the parabola shifted $7$ units to the right multiplied by $1/4$, so it's strictly decreasing on $a\le 7$. Therefore, to maximise $bc$, in the first inequality, we want to set $\dfrac{(7-a)^2}{4}$ as large as possible, and make it an equality, which is equivalent to $b=c$. This gives the solution $a = 3$, $b = c = 2$, so $bc = 4$.
$$ a^2=b^2+c^2+\frac{bc}{4}$$ so $$ (7-b-c)^2 = b^2+c^2+bc/4$$ so $$ 49-14b-14c+2bc = bc/4$$ so $$ 7bc= 56(b+c)-196 \geq 112\sqrt{bc}-196$$ Put $ x = \sqrt{bc} \;\;\;\;(\leq {b+c\over 2} < {7\over 2})$ then $$7x^2-112x+196\geq 0$$ so $$x^2-16x+28\geq 0$$ so $$x\in (0,2]\cup[14,\infty)$$ and thus $x_{\max} \leq 2$ (and this is achieved if $b=c=2$)