, 2006

(by Riddhipratim Basu) We introduce $s - a = x$, $s - b = y$, $s - c = z$, where $s = (a + b + c)/2$. Then $x$, $y$, $z$ are positive, and $a = y + z$, $b = z + x$, $c = x + y$. We may express
$$\frac{R}{2r} = \frac{abcs}{4\Delta^2} = \frac{(x+y)(y+z)(z+x)}{8xyz}.$$
Similarly
$$\begin{align*} 4a^2 - (b-c)^2 &= (3y+z)(3z+y), & 4b^2 - (c-a)^2 &= (3x+z)(3z+x), \\ 4c^2 - (a-b)^2 &= (3x+y)(3y+x). \end{align*}$$
Thus we need to prove
$$\frac{(x+y)(y+z)(z+x)}{8xyz} \ge \frac{64^2(x+y)^4(y+z)^4(z+x)^4}{(3x+y)^2(3y+x)^2(3y+z)^2(3z+y)^2(3x+z)^2(3z+x)^2}$$
This may be written in the form
$$(3x+y)^2(3y+x)^2(3y+z)^2(3z+y)^2(3x+z)^2(3z+x)^2 \geq 32^3 xyz (x+y)^3 (y+z)^4 (z+x)^3$$
Using AM-GM inequality, we have
$$\text{(3x + y)(3y + x) = 3x 2 + 3y 2 + 10xy = (x + y) 2 + (x + y) 2 + (x + y) 2 + 4xy 4 (x + y) 6 4xy 1/4 .}$$
It follows that $(3x+y)^2(3y+x)^2 \geq 32\sqrt{xy}(x+y)^3$ and similar expressions
for the other products. Thus
$$(3x+y)^2(3y+x)^2(3y+z)^2(3z+y)^2(3x+z)^2(3z+x)^2 \geq 32^3 xyz (x+y)^3 (y+z)^4 (z+x)^3,$$