<s>[INST] Let a,b and c be the side lengths of triangle ABC respectively...find the greatest value of b*c. 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:
$$a+b+c=7 \implies b+c=7-a,\quad \cos A=-\frac{1}{8}\\
a^2=b^2+c^2-2bc\cdot \cos A\\
\implies a^2=b^2+c^2+\frac{bc}{4} \\
\implies a^2=(7-a)^2-2bc+\frac{bc}{4} \\
\implies bc=4(7-2a)$$
 [/INST] 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 $bc = 4$.
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