Consider a right angle triangle with sides $a , b, c$ with the inradius of $r$ and circumradius of $R$. Let $S$ be the area of the right angle triangle . Show that
$r+R > \sqrt{2S}$
I had tried my way by using Cauchy’s theorem which resulted $r+R>\sqrt{\frac{abc}{p}}$. And I don’t know what should I do next .! Pls can anyone help me !
with $$r=\frac{a+b-c}{2}$$ and $$R=\frac{c}{2}$$ and $$S=\frac{ab}{2}$$ we get $$\frac{a+b}{2}\geq \sqrt{ab}$$ which is true. it must be $$R+r\geq \sqrt{2S}$$