arqady wrote: Let $P$ be a point inside a triangle $ABC$, $d_a$, $d_b$ and $d_c$ be distances from $P$ to the lines $BC$, $AC$ and $AB$ respectively, $R$ be a radius of the circumcircle and $r$ be a radius of the inscribed circle for $\Delta ABC.$ Prove that: $$\sqrt{d_a}+\sqrt{d_b}+\sqrt{d_c}\leq\sqrt{2R+5r}.$$ Looks nice Here's my solution: Note that $ad_a+bd_b+cd_c=2S$, therefore by Cauchy-Schwarz inequality: $$\sqrt{d_a}+\sqrt{d_b}+\sqrt{d_c}\leq \sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\cdot (ad_a+bd_b+cd_c)}=\sqrt{\dfrac{ab+bc+ca}{abc}\cdot 2S}=\sqrt{\dfrac{p^2+r^2+4Rr}{2R}}$$Thus, it remains to prove that: $\dfrac{p^2+r^2+4Rr}{2R}\leq 2R+5r$, or: $p^2\leq 4R^2+6Rr-r^2$. This is true by Gerretsen's inequality: $$p^2\leq 4R^2+4Rr+3r^2=4R^2+6Rr-r^2-2r(R-2r)\leq 4R^2+6Rr-r^2$$

An excellent proof, an excellent problem.

And the official solution is Ravi-substitution bash..

shalomrav wrote: And the official solution is Ravi-substitution bash.. Are you sure?

arqady wrote: shalomrav wrote: And the official solution is Ravi-substitution bash.. Are you sure? After the standard Qauchy, yes. At least that's the solution on the Gillis website... Is there another solution?