Prove the inequality $R+r > \sqrt{2S}$

Question

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 !

This is a hint for tackling a more general problem: by Carnot's theorem, $R+r$ is the sum of the distances of the circumcenter from the triangle sides. — Jack D'Aurizio, Nov 11 '17 at 15:20

score 1 · Accepted Answer · answered Nov 11 '17 at 14:55

1

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 $$R+r\geq \sqrt{2S}$$

answered Nov 11 '17 at 14:55

Dr. Sonnhard Graubner

95,283

Is $$r= \frac{a+b-c}{2} $$ getting from $r=p-c$ because the circle is inscribed in the right angle triangle ? – November ft Blue Nov 11 '17 at 15:10
yes this is the incircle radius – Dr. Sonnhard Graubner Nov 11 '17 at 15:15
I was thinking too far! I used the formula $S= \frac{abc}{4R}$ and $S=pr$ which made the situation harder! Anyway thank you so much. – November ft Blue Nov 11 '17 at 15:18
thank you for this nice problem – Dr. Sonnhard Graubner Nov 11 '17 at 15:36

Prove the inequality $R+r > \sqrt{2S}$

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