How can i prove this inequality

Question

If c is the hypotenuse of a right triangle with legs $a,b$ and radius $r$ of the incircle of this triangle then prove $$c/r ≥ 2(\sqrt{2}+1)$$ How can i solve this? please help :/

Answer 1

Hint:

Note that the the maximum value of $r$ is given when the triangle is isosceles with $a=b=c/\sqrt{2}$ and, in this case we have $$ r=a-\frac{1}{2}c=\frac{c(\sqrt{2}-1)}{2} $$

and find $c/r$

Answer 2

Note that $ (ABC) = \frac{(a+b+c)r}{2} $. Since $\triangle ABC $ is right, it follows that $ (ABC) = \frac{ab}{2} $.

So we conclude that $ r = \frac{ab}{a+b+c}$. Now observe that $$ \frac{c}{r} = \frac{(a+b+c)c}{ab} = \frac{(a+b)\sqrt{a^2+b^2} + a^2 + b^2}{ab} \geq \frac{2\sqrt{ab}\cdot \sqrt{2ab} + 2ab}{ab} =2(\sqrt{2} +1) $$

Answer 3

Assume your triangle has sides $\{a,b,c\}$.
Remark that $$r=\frac {ab}{a+b+c}$$ (from the usual formula connecting inradius to area and semiperimeter). It follows that your inequality is equivalent to $$2\left(\sqrt 2+1\right)ab≤c\left(a+b+c\right)$$ Now, we have $$c^2=a^2+b^2≥2ab\implies c≥\sqrt {2ab}$$

And we have the usual arithmetic-geometric inequality $$\frac {a+b}2≥\sqrt {ab}$$ It follows that $$c\left(a+b+c\right)≥\sqrt {2ab}\left(2\sqrt {ab}+\sqrt {2ab}\right)=2\sqrt2\,ab+2ab$$ And this is the desired inequality.