Question

If a, b, c are the sides of a triangle and s = $\frac{a+b+c}{2}$, then prove that 8(s- a)(s-b)(s-c) $\le$ abc.

Answer 1

Given,

$8(s-a)(s-b)(s-c)\le abc$

$\frac{abc}{8(s-a)(s-b)(s-c)}\ge 1$

$\frac{abc}{8}\ge (s-a)(s-b)(s-c)$

take $s-a=x\ge 0,s-b=y\ge 0,s-c=z\ge 0$

Hence $(s-a)+(s-b)=x+y$

$\Rightarrow 2x-a-b=x+y$

$\therefore c=x+y$

Similarly, we get, $a=y+z,b=x+z$

$\frac{x+y}{2}\cdot \frac{z+y}{2}\cdot \frac{x+z}{2}\ge xyz$ which is true.

$\therefore 8(s-a)(s-b)(s-c)\le abc$