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:
$$a+b+c=7 \implies b+c=7-a,\quad \cos A=-\frac{1}{8}\\
a^2=b^2+c^2-2bc\cdot \cos A\\
\implies a^2=b^2+c^2+\frac{bc}{4} \\
\implies a^2=(7-a)^2-2bc+\frac{bc}{4} \\
\implies bc=4(7-2a)$$

$$ a^2=b^2+c^2+\frac{bc}{4}$$
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 $b=c=2$)
