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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.

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

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