Question

In triangle ABC, prove that
$\sum \frac{co{s}^{2}A+co{s}^{2}B}{cosA+B}\ge 1+\frac{r}{R}$

Answer 1

Let cos A = x, cos B = y, cos C = z then,
x + y = 2 cos $\frac{A+B}{2}$ cos $\frac{A-B}{2}$ > 0
$x^2+y^2\ge 2xy$
Add $x^2+y^2$ to both sides
$\therefore 2(x^2+y^2)>(x+y{)}^2$
Since x + y is positive we can divide by it
$\therefore \frac{2(x^2+y^2)}{(x+y)}>x+y$
Similarly write other inequalities and add
$\therefore 2\sum \frac{(x^2+y^2)}{(x+y)}>2(x+y+z)$
$or\sum \frac{(co{s}^{2}A+co{s}^B)}{(cosA+cosB)}>(cosA+cosB+cosC)>1+\frac{r}{R}$