Let $ABC$ be an acute triangle, and $H$ its orthocenter. The distance from $H$ to rays $BC$, $CA$, and $AB$ is denoted by $d_a$, $d_b$, and $d_c$, respectively. Let $R$ be the radius of circumcenter of $\triangle ABC$ and $r$ be the radius of incenter of $\triangle ABC$. Prove the following inequality: $$d_a+d_b+d_c \le \frac{3R^2}{4r}$$.
Problem
Source: 2017 Moldova TST, Day 2, Problem 3
Tags: geometric inequality, inequalities, geometry, circumcircle, incenter
Snakes
19.03.2017 18:38
Firstly we use Euler's inequality ($R\geq 2r$) and turn our inequality into: $d_{a}+d_{b}+d_{c}\leq \frac{3}{2}R$
Now using Erdős–Mordell inequality we have $d_{a}+d_{b}+d_{c}\leq \frac{HA+HB+HC}{2}$
But $HA=2Rcos A$ so we only have to prove that $cos A +cos B+ cos C\leq \frac{3}{2}$ and this can be easily proved using JensenTake $f(x)=cos x$, the $f^{''}(x)=-cos x<o$ for all $x\in \left(o, \frac{\pi}{2}\right)$ thus our function is concave as the triangle is acute.
Finally, by Jensen's Inequality
$$cos A+cos B+cos C\leq 3cos \left(\frac{A+B+C}{3}\right)=3 cos 60=\frac{3}{2}$$
andrei.pantea
19.03.2017 18:49
Snakes wrote:
Firstly we use Euler's inequality ($R\geq 2r$) and turn our inequality into: $d_{a}+d_{b}+d_{c}\leq \frac{3}{2}R$
Now using Erdős–Mordell inequality we have $d_{a}+d_{b}+d_{c}\leq \frac{HA+HB+HC}{2}$
But $HA=2Rcos A$ so we only have to prove that $cos A +cos B+ cos C\leq \frac{3}{2}$ and this can be easily proved using JensenTake $f(x)=cos x$, the $f^{''}(x)=-cos x<o$ for all $x\in \left(o, \frac{\pi}{2}\right)$ thus our function is concave as the triangle is acute.
Finally, by Jensen's Inequality
$$cos A+cos B+cos C\leq 3cos \left(\frac{A+B+C}{3}\right)=3 cos 60=\frac{3}{2}$$
Another way is by Carnot's Relation: $$cosA+cosB+cosC=1+\frac{r}{R}$$