# Prove for a right triangle $Ac^{-1} \geq \max(a^{-1},b^{-1})$

Given a right triangle with the legs $a$ and $b$ and hypotenuse $c$, I need to show that $\exists$A - const, s.t.

$Ac^{-1} \geq \max(a^{-1},b^{-1})$

and express $A$ in term of $\gamma = \dfrac{c}{r}$, where $r$ is a radius of an inscribed circle.

I used the fact that $r = \dfrac{a + b - c}{2}$ to get

$\dfrac{1}{(2\gamma + 1)}c^{-1} = \dfrac{1}{a + b} \leq \dfrac{1}{a} + \dfrac{1}{b} \leq 2\max(a^{-1}, b^{-1}),$

$\dfrac{1}{2(2\gamma + 1)}c^{-1} \leq \max(a^{-1}, b^{-1})$,

which is the opposite of what I need to show. Is there a way to get the required inequality from here or do I need to proceed in a completely different way?

We may suppose that $a\ge b$.
Using that $$R=\frac c2$$ $$\frac Rr\ge \frac bc+\frac cb\tag1$$ (The proof for $(1)$ can be seen here)
we have $$\frac Rr=\frac{\frac c2}{r}\ge\frac bc+\frac cb\ge\frac cb$$ from which $$\frac{\gamma}{2}\cdot c^{-1}\ge\max(a^{-1},b^{-1})$$ follows.
You just need to find an inequality the other way around. Do you think you can argue why $\dfrac{1}{a + b} \geq \delta \max(a^{-1}, b^{-1})$ is true for some $\delta > 0$?
The notation could be simplified by saying "After renaming I assume WLOG that $b > a$, therefore $\max(a^{-1}, b^{-1}) = a^{-1}$"
• Existence of such $\delta$ is pretty obvious, but how does it help me to express the constant in terms of $\gamma$? – Moisej Braver Nov 16 '17 at 12:57