# Proving $R +r\le h_{max}$

If $R$ is the circumradius , $r$ the inradius and $h_{max}$ is the largest altitude of acute angled triangle $ABC$, then prove that $$R +r\le h_{max}.$$ I tried this using Euler's inequality but I did not succeed.

• I forgot to mention that triangle ABC is acute. Does it affect the answer – Achal Kumar Oct 11 '14 at 14:22
• it is relevant in order to make my $(2)$ work. Have also a look at gogeometry.com/geometry/… – Jack D'Aurizio Oct 11 '14 at 14:45

By assuming $a\leq b\leq c$ we have that the greatest altitude is $h_a$ and:
$$R = \frac{abc}{4\Delta},\quad r=\frac{2\Delta}{a+b+c},\quad h_a=\frac{2\Delta}{a}$$ so, by Heron's formula, we have to prove that:
$$2abc + (-a+b+c)(a-b+c)(a+b-c) \leq \frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{a}$$ or: $$2a^2 bc\leq(b+c)(-a+b+c)(a-b+c)(a+b-c)\tag{1}$$ where $2a^2\leq(b+c)(b+c-a)$ is trivial (since $c\geq b\geq a$) and $$bc \leq (a-b+c)(a+b-c) = a^2-(b-c)^2$$ is equivalent to: $$b^2+c^2-bc \leq a^2 \tag{2}$$ that follows from the cosine theorem, since $\widehat{A}\leq\frac{\pi}{3}$ (otherwise, $a$ cannot be the shortest side).